Polynomial maps with nilpotent Jacobians in dimension three II

Dan Yan 111 The author are supported by the Natural Science Foundation of Hunan Province (Grant No.2016JJ3085), the National Natural Science Foundation of China (Grant No.11601146) and the Construct Program of the Key Discipline in Hunan Province.
Key Laboratory of HPCSIP,
College of Mathematics and Computer Science,
Hunan Normal University, Changsha 410081, China
E-mail: yan-dan-hi@163.com
Abstract

In the paper, we first classify all polynomial maps of the form H=(u​(x,y),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y),v(x,y,z),h(x,y)) in the case that J​H𝐽𝐻JH is nilpotent and (degy⁑u,degy⁑h)≀3subscriptdegree𝑦𝑒subscriptdegreeπ‘¦β„Ž3(\deg_{y}u,\deg_{y}h)\leq 3, H​(0)=0𝐻00H(0)=0. Then we classify all polynomial maps of the form H=(u​(x,y,z),v​(x,y,u),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦π‘’β„Žπ‘₯𝑦H=(u(x,y,z),v(x,y,u),h(x,y)) in the case that J​H𝐽𝐻JH is nilpotent and (deg⁑v​(x,y,0),deg⁑h)≀3degree𝑣π‘₯𝑦0degreeβ„Ž3(\deg v(x,y,0),\allowbreak\deg h)\leq 3, H​(0)=0𝐻00H(0)=0. Finally, we classify polynomial maps of the form H=(u​(x,y,z),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y,z),v(x,y,z),h(x,y)) in certain conditions.

Keywords. Jacobian Conjecture, Nilpotent Jacobian matrix, Polynomial maps
MSC(2010). Primary 14E05; Secondary 14A05;14R15

1 Introduction

Throughout this paper, we will write 𝐊𝐊{\bf K} for algebraically closed field and πŠβ€‹[X]=πŠβ€‹[x1,x2,…,xn]𝐊delimited-[]π‘‹πŠsubscriptπ‘₯1subscriptπ‘₯2…subscriptπ‘₯𝑛{\bf K}[X]={\bf K}[x_{1},x_{2},\ldots,x_{n}] (πŠβ€‹[XΒ―]=πŠβ€‹[x,y,z]𝐊delimited-[]Β―π‘‹πŠπ‘₯𝑦𝑧{\bf K}[\bar{X}]={\bf K}[x,y,z])for the polynomial algebra over 𝐊𝐊{\bf K} with n𝑛n (333) indeterminates. Let F=(F1,F2,…,Fn):𝐊nβ†’πŠn:𝐹subscript𝐹1subscript𝐹2…subscript𝐹𝑛→superscriptπŠπ‘›superscriptπŠπ‘›F=(F_{1},F_{2},\ldots,F_{n}):{\bf{K}}^{n}\rightarrow{\bf{K}}^{n} be a polynomial map, that is, FiβˆˆπŠβ€‹[X]subscriptπΉπ‘–πŠdelimited-[]𝑋F_{i}\in{\bf{K}}[X] for all 1≀i≀n1𝑖𝑛1\leq i\leq n. Let J​F=(βˆ‚Fiβˆ‚xj)nΓ—n𝐽𝐹subscriptsubscript𝐹𝑖subscriptπ‘₯𝑗𝑛𝑛JF=(\frac{\partial F_{i}}{\partial x_{j}})_{n\times n} be the Jacobian matrix of F𝐹F.

The Jacobian Conjecture (JC) raised by O.H. Keller in 1939 in [8] states that a polynomial map F:𝐊nβ†’πŠn:𝐹→superscriptπŠπ‘›superscriptπŠπ‘›F:{\bf{K}}^{n}\rightarrow{\bf{K}}^{n} is invertible if the Jacobian determinant detJ​F𝐽𝐹\det JF is a nonzero constant. This conjecture has been attacked by many people from various research fields, but it is still open, even for nβ‰₯2𝑛2n\geq 2. Only the case n=1𝑛1n=1 is obvious. For more information about the wonderful 70-year history, see [1], [5], and the references therein.

In 1980, S.S.S.Wang ([9]) showed that the JC holds for all polynomial maps of degree 2 in all dimensions (up to an affine transformation). The most powerful result is the reduction to degree 3, due to H.Bass, E.Connell and D.Wright ([1]) in 1982 and A.Yagzhev ([11]) in 1980, which asserts that the JC is true if the JC holds for all polynomial maps X+H𝑋𝐻X+H, where H𝐻H is homogeneous of degree 3. Thus, many authors study these maps and led to pose the following problem.

(Homogeneous) dependence problem. Let H=(H1,…,Hn)βˆˆπŠβ€‹[X]𝐻subscript𝐻1…subscriptπ»π‘›πŠdelimited-[]𝑋H=(H_{1},\ldots,H_{n})\in{\bf K}[X] be a (homogeneous) polynomial map of degree d𝑑d such that J​H𝐽𝐻JH is nilpotent and H​(0)=0𝐻00H(0)=0. Whether H1,…,Hnsubscript𝐻1…subscript𝐻𝑛H_{1},\ldots,H_{n} are linearly dependent over 𝐊𝐊{\bf K}?

The answer to the above problem is affirmative if rankJ​H≀1𝐽𝐻1JH\leq 1 ([1]). In particular, this implies that the dependence problem has an affirmative answer in the case n=2𝑛2n=2. D. Wright give an affirmative answer when H𝐻H is homogeneous of degree 3 in the case n=3𝑛3n=3 ([10]) and the case n=4𝑛4n=4 is solved by Hubbers in [7]. M. de Bondt and A. van den Essen give an affirmative answer to the above problem in the case H𝐻H is homogeneous and n=3𝑛3n=3 ([3]). A. van den Essen finds the first counterexample in dimension three for the dependence problem ([6]). M. de Bondt give a negative answer to the homogeneous dependence problem for dβ‰₯3𝑑3d\geq 3. In particular, he constructed counterexamples to the problem for all dimensions nβ‰₯5𝑛5n\geq 5 ([2]). In [4], M. Chamberland and A. van den Essen classify all polynomial maps of the form H=(u​(x,y),v​(x,y,z),h​(u​(x,y),v​(x,y,z)))𝐻𝑒π‘₯𝑦𝑣π‘₯π‘¦π‘§β„Žπ‘’π‘₯𝑦𝑣π‘₯𝑦𝑧H=(u(x,y),v(x,y,z),h(u(x,y),v(x,y,z))) with J​H𝐽𝐻JH nilpotent. In particular, they show that all maps of this form with H​(0)=0𝐻00H(0)=0, J​H𝐽𝐻JH nilpotent and H1,H2,H3subscript𝐻1subscript𝐻2subscript𝐻3H_{1},H_{2},H_{3} are linearly independent has the same form as the counterexample that gave by A. van den Essen in [6] (up to a linear coordinate change). We classify all polynomial maps of the form H=(u​(x,y),v​(x,y,z),h​(x,y,z))𝐻𝑒π‘₯𝑦𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦𝑧H=(u(x,y),v(x,y,z),h(x,y,z)) in the case that J​H𝐽𝐻JH is nilpotent and degz⁑v≀3subscriptdegree𝑧𝑣3\deg_{z}v\leq 3, (degy⁑u​(x,y),degy⁑h​(x,y,z))=1subscriptdegree𝑦𝑒π‘₯𝑦subscriptdegreeπ‘¦β„Žπ‘₯𝑦𝑧1(\deg_{y}u(x,y),\deg_{y}h(x,y,z))=1 ([12]) and classify all polynomial maps of the form H=(u​(x,y),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y),v(x,y,z),h(x,y)) in the case that J​H𝐽𝐻JH is nilpotent and (degy⁑u​(x,y),degy⁑h​(x,y))≀2subscriptdegree𝑦𝑒π‘₯𝑦subscriptdegreeπ‘¦β„Žπ‘₯𝑦2(\deg_{y}u(x,y),\deg_{y}h(x,y))\leq 2 or degy⁑u​(x,y)subscriptdegree𝑦𝑒π‘₯𝑦\deg_{y}u(x,y) or degy⁑h​(x,y)subscriptdegreeπ‘¦β„Žπ‘₯𝑦\deg_{y}h(x,y) is a prime number ([13]).

In section 2, we classify all polynomial maps of the form H=(u​(x,y),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y),v(x,y,z),\allowbreak h(x,y)) in the case that J​H𝐽𝐻JH is nilpotent and (degy⁑u,degy⁑h)≀3subscriptdegree𝑦𝑒subscriptdegreeπ‘¦β„Ž3(\deg_{y}u,\deg_{y}h)\leq 3, H​(0)=0𝐻00H(0)=0. Then, in section 3, we classify all polynomial maps of the form H=(u​(x,y,z),v​(x,y,u),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦π‘’β„Žπ‘₯𝑦H=(u(x,y,z),v(x,y,u),\allowbreak h(x,y)) in the case that J​H𝐽𝐻JH is nilpotent and (deg⁑v​(x,y,0),deg⁑h)≀3degree𝑣π‘₯𝑦0degreeβ„Ž3(\deg v(x,y,0),\deg h)\leq 3, H​(0)=0𝐻00H(0)=0. We prove that u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent in the case that J​H𝐽𝐻JH is nilpotent and H𝐻H has the form: H=(u​(x,y,z),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y,z),v(x,y,z),h(x,y)) with degz⁑v=1subscriptdegree𝑧𝑣1\deg_{z}v=1 and degz⁑uβ‰₯2subscriptdegree𝑧𝑒2\deg_{z}u\geq 2, H​(0)=0𝐻00H(0)=0 in section 4. The main results in the paper are Theorem 2.2, Theorem 2.5, Theorem 3.2 and Theorem 3.3, Theorem 4.1. We define Qxi:=βˆ‚Qβˆ‚xiassignsubscript𝑄subscriptπ‘₯𝑖𝑄subscriptπ‘₯𝑖Q_{x_{i}}:=\frac{\partial Q}{\partial x_{i}} and that degy⁑fsubscriptdegree𝑦𝑓\deg_{y}f is the highest degree of y𝑦y in f𝑓f.

2 Polynomial maps of the form H=(u​(x,y),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y),v(x,\allowbreak y,z),h(x,y))

In this section, we classify all polynomial maps of the form H=(u​(x,y),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y),v(x,y,z),\allowbreak h(x,y)) in the case that J​H𝐽𝐻JH is nilpotent and (degy⁑u,degy⁑h)≀3subscriptdegree𝑦𝑒subscriptdegreeπ‘¦β„Ž3(\deg_{y}u,\deg_{y}h)\leq 3, H​(0)=0𝐻00H(0)=0.

Lemma 2.1.

Let QβˆˆπŠβ€‹[x,y]π‘„πŠπ‘₯𝑦Q\in{\bf K}[x,y] and Q​(x,y)=Qr​(x)​yr+Qrβˆ’1​(x)​yrβˆ’1+β‹―+Q1​(x)​y+Q0​(x)𝑄π‘₯𝑦subscriptπ‘„π‘Ÿπ‘₯superscriptπ‘¦π‘Ÿsubscriptπ‘„π‘Ÿ1π‘₯superscriptπ‘¦π‘Ÿ1β‹―subscript𝑄1π‘₯𝑦subscript𝑄0π‘₯Q(x,y)=Q_{r}(x)y^{r}+Q_{r-1}(x)y^{r-1}+\cdots+Q_{1}(x)y+Q_{0}(x) with Qr​(x)βˆˆπŠβˆ—subscriptπ‘„π‘Ÿπ‘₯superscript𝐊Q_{r}(x)\in{\bf K}^{*}. If Qy|Qxconditionalsubscript𝑄𝑦subscript𝑄π‘₯Q_{y}|Q_{x} or Qy|Q+cconditionalsubscript𝑄𝑦𝑄𝑐Q_{y}|Q+c for some cβˆˆπŠπ‘πŠc\in{\bf K}, then Q𝑄Q is a polynomial of y+a​(x)π‘¦π‘Žπ‘₯y+a(x) for some a​(x)βˆˆπŠβ€‹[x]π‘Žπ‘₯𝐊delimited-[]π‘₯a(x)\in{\bf K}[x].

Proof.

Let Qr:=Qr​(x)assignsubscriptπ‘„π‘Ÿsubscriptπ‘„π‘Ÿπ‘₯Q_{r}:=Q_{r}(x). Then we have Qy=r​Qr​yrβˆ’1+(rβˆ’1)​Qrβˆ’1​(x)​yrβˆ’2+β‹―+Q1​(x)subscriptπ‘„π‘¦π‘Ÿsubscriptπ‘„π‘Ÿsuperscriptπ‘¦π‘Ÿ1π‘Ÿ1subscriptπ‘„π‘Ÿ1π‘₯superscriptπ‘¦π‘Ÿ2β‹―subscript𝑄1π‘₯Q_{y}=rQ_{r}y^{r-1}+(r-1)Q_{r-1}(x)y^{r-2}+\cdots+Q_{1}(x) and Qx=Qrβˆ’1′​(x)​yrβˆ’1+β‹―+Q1′​(x)​y+Q0′​(x)subscript𝑄π‘₯superscriptsubscriptπ‘„π‘Ÿ1β€²π‘₯superscriptπ‘¦π‘Ÿ1β‹―superscriptsubscript𝑄1β€²π‘₯𝑦superscriptsubscript𝑄0β€²π‘₯Q_{x}=Q_{r-1}^{\prime}(x)y^{r-1}+\cdots+Q_{1}^{\prime}(x)y+Q_{0}^{\prime}(x). Thus, we have degy⁑Qyβ‰₯degy⁑Qxsubscriptdegree𝑦subscript𝑄𝑦subscriptdegree𝑦subscript𝑄π‘₯\deg_{y}Q_{y}\geq\deg_{y}Q_{x}. We always view that the polynomials are in πŠβ€‹[x]​[y]𝐊delimited-[]π‘₯delimited-[]𝑦{\bf K}[x][y] with coefficients in πŠβ€‹[x]𝐊delimited-[]π‘₯{\bf K}[x] when comparing the coefficients of yisuperscript𝑦𝑖y^{i}.

Case I If Qy|Qxconditionalsubscript𝑄𝑦subscript𝑄π‘₯Q_{y}|Q_{x}, then we have

Qx=b​(x)​Qy(2.1)subscript𝑄π‘₯𝑏π‘₯subscript𝑄𝑦2.1Q_{x}=b(x)Q_{y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.1)

for some b​(x)βˆˆπŠβ€‹[x]𝑏π‘₯𝐊delimited-[]π‘₯b(x)\in{\bf K}[x]. That is,

Qrβˆ’1′​(x)​yrβˆ’1+β‹―+Q1′​(x)​y+Q0′​(x)=b​(x)​(r​Qr​yrβˆ’1+β‹―+Q1​(x))superscriptsubscriptπ‘„π‘Ÿ1β€²π‘₯superscriptπ‘¦π‘Ÿ1β‹―superscriptsubscript𝑄1β€²π‘₯𝑦superscriptsubscript𝑄0β€²π‘₯𝑏π‘₯π‘Ÿsubscriptπ‘„π‘Ÿsuperscriptπ‘¦π‘Ÿ1β‹―subscript𝑄1π‘₯Q_{r-1}^{\prime}(x)y^{r-1}+\cdots+Q_{1}^{\prime}(x)y+Q_{0}^{\prime}(x)=b(x)(rQ_{r}y^{r-1}+\cdots+Q_{1}(x))

Comparing the coefficients of yrβˆ’1superscriptπ‘¦π‘Ÿ1y^{r-1} of the above equation, we have that b​(x)=Qrβˆ’1′​(x)r​Qr𝑏π‘₯superscriptsubscriptπ‘„π‘Ÿ1β€²π‘₯π‘Ÿsubscriptπ‘„π‘Ÿb(x)=\frac{Q_{r-1}^{\prime}(x)}{rQ_{r}}. Then let xΒ―=xΒ―π‘₯π‘₯\bar{x}=x, yΒ―=y+Qrβˆ’1​(x)r​Qr¯𝑦𝑦subscriptπ‘„π‘Ÿ1π‘₯π‘Ÿsubscriptπ‘„π‘Ÿ\bar{y}=y+\frac{Q_{r-1}(x)}{rQ_{r}}, it follows from equation (2.1)2.1(2.1) that QxΒ―=0subscript𝑄¯π‘₯0Q_{\bar{x}}=0. That is, Q​(x,y)βˆˆπŠβ€‹[y+Qrβˆ’1​(x)r​Qr]𝑄π‘₯π‘¦πŠdelimited-[]𝑦subscriptπ‘„π‘Ÿ1π‘₯π‘Ÿsubscriptπ‘„π‘ŸQ(x,y)\in{\bf K}[y+\frac{Q_{r-1}(x)}{rQ_{r}}]. Let a​(x)=Qrβˆ’1​(x)r​Qrπ‘Žπ‘₯subscriptπ‘„π‘Ÿ1π‘₯π‘Ÿsubscriptπ‘„π‘Ÿa(x)=\frac{Q_{r-1}(x)}{rQ_{r}}. Then the conclusion follows.

Case II If Qy|(Q+c)conditionalsubscript𝑄𝑦𝑄𝑐Q_{y}|(Q+c), then Q+c=(e​(x)​y+d​(x))​Qy𝑄𝑐𝑒π‘₯𝑦𝑑π‘₯subscript𝑄𝑦Q+c=(e(x)y+d(x))Q_{y}. That is,
Qr​yr+Qrβˆ’1​(x)​yrβˆ’1+β‹―+Q1​(x)​y+Q0​(x)+c=(e​(x)​y+d​(x))​(r​Qr​yrβˆ’1+(rβˆ’1)​Qrβˆ’1​(x)​yrβˆ’2+β‹―+Q1​(x))subscriptπ‘„π‘Ÿsuperscriptπ‘¦π‘Ÿsubscriptπ‘„π‘Ÿ1π‘₯superscriptπ‘¦π‘Ÿ1β‹―subscript𝑄1π‘₯𝑦subscript𝑄0π‘₯𝑐𝑒π‘₯𝑦𝑑π‘₯π‘Ÿsubscriptπ‘„π‘Ÿsuperscriptπ‘¦π‘Ÿ1π‘Ÿ1subscriptπ‘„π‘Ÿ1π‘₯superscriptπ‘¦π‘Ÿ2β‹―subscript𝑄1π‘₯Q_{r}y^{r}+Q_{r-1}(x)y^{r-1}+\cdots+Q_{1}(x)y+Q_{0}(x)+c=(e(x)y+d(x))(rQ_{r}y^{r-1}+(r-1)Q_{r-1}(x)y^{r-2}+\cdots+Q_{1}(x))Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (2.2)
Comparing the coefficients of yrsuperscriptπ‘¦π‘Ÿy^{r} and yrβˆ’1superscriptπ‘¦π‘Ÿ1y^{r-1} of equation (2.2)2.2(2.2), we have that e​(x)=1r𝑒π‘₯1π‘Ÿe(x)=\frac{1}{r} and d​(x)=Qrβˆ’1​(x)r2​Qr𝑑π‘₯subscriptπ‘„π‘Ÿ1π‘₯superscriptπ‘Ÿ2subscriptπ‘„π‘Ÿd(x)=\frac{Q_{r-1}(x)}{r^{2}Q_{r}}. Then equation (2.2)2.2(2.2) has the following form:
Qr​yr+Qrβˆ’1​(x)​yrβˆ’1+β‹―+Q1​(x)​y+Q0​(x)+c=(1r​y+Qrβˆ’1​(x)r2​Qr)​(r​Qr​yrβˆ’1+(rβˆ’1)​Qrβˆ’1​(x)​yrβˆ’2+β‹―+Q1​(x))subscriptπ‘„π‘Ÿsuperscriptπ‘¦π‘Ÿsubscriptπ‘„π‘Ÿ1π‘₯superscriptπ‘¦π‘Ÿ1β‹―subscript𝑄1π‘₯𝑦subscript𝑄0π‘₯𝑐1π‘Ÿπ‘¦subscriptπ‘„π‘Ÿ1π‘₯superscriptπ‘Ÿ2subscriptπ‘„π‘Ÿπ‘Ÿsubscriptπ‘„π‘Ÿsuperscriptπ‘¦π‘Ÿ1π‘Ÿ1subscriptπ‘„π‘Ÿ1π‘₯superscriptπ‘¦π‘Ÿ2β‹―subscript𝑄1π‘₯Q_{r}y^{r}+Q_{r-1}(x)y^{r-1}+\cdots+Q_{1}(x)y+Q_{0}(x)+c=(\frac{1}{r}y+\frac{Q_{r-1}(x)}{r^{2}Q_{r}})(rQ_{r}y^{r-1}+(r-1)Q_{r-1}(x)y^{r-2}+\cdots+Q_{1}(x))Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (2.3)

Claim: Qi​(x)=Cri​Qrβˆ’1rβˆ’i​(x)rrβˆ’i​Qrrβˆ’iβˆ’1​(x)subscript𝑄𝑖π‘₯superscriptsubscriptπΆπ‘Ÿπ‘–superscriptsubscriptπ‘„π‘Ÿ1π‘Ÿπ‘–π‘₯superscriptπ‘Ÿπ‘Ÿπ‘–superscriptsubscriptπ‘„π‘Ÿπ‘Ÿπ‘–1π‘₯Q_{i}(x)=C_{r}^{i}\frac{Q_{r-1}^{r-i}(x)}{r^{r-i}Q_{r}^{r-i-1}(x)} for 1≀i≀rβˆ’21π‘–π‘Ÿ21\leq i\leq r-2 and Cri=r!i!​(rβˆ’i)!superscriptsubscriptπΆπ‘Ÿπ‘–π‘Ÿπ‘–π‘Ÿπ‘–C_{r}^{i}=\frac{r!}{i!(r-i)!}.

Comparing the coefficients of yrβˆ’2superscriptπ‘¦π‘Ÿ2y^{r-2} of equation (2.3)2.3(2.3), we have the following equation:

Qrβˆ’2​(x)=Crrβˆ’2​Qrβˆ’12​(x)r2​Qrsubscriptπ‘„π‘Ÿ2π‘₯superscriptsubscriptπΆπ‘Ÿπ‘Ÿ2superscriptsubscriptπ‘„π‘Ÿ12π‘₯superscriptπ‘Ÿ2subscriptπ‘„π‘ŸQ_{r-2}(x)=C_{r}^{r-2}\frac{Q_{r-1}^{2}(x)}{r^{2}Q_{r}}

Suppose Qk+1​(x)=Crk+1​Qrβˆ’1rβˆ’kβˆ’1​(x)rrβˆ’kβˆ’1​Qrrβˆ’kβˆ’2​(x)subscriptπ‘„π‘˜1π‘₯superscriptsubscriptπΆπ‘Ÿπ‘˜1superscriptsubscriptπ‘„π‘Ÿ1π‘Ÿπ‘˜1π‘₯superscriptπ‘Ÿπ‘Ÿπ‘˜1superscriptsubscriptπ‘„π‘Ÿπ‘Ÿπ‘˜2π‘₯Q_{k+1}(x)=C_{r}^{k+1}\frac{Q_{r-1}^{r-k-1}(x)}{r^{r-k-1}Q_{r}^{r-k-2}(x)}. Then comparing the coefficients of yksuperscriptπ‘¦π‘˜y^{k} of equation (2.3)2.3(2.3), we have the following equation:

Qk​(x)=1r​k​Qk​(x)+Qrβˆ’1​(x)r2​Qr​(k+1)​Qk+1​(x).subscriptπ‘„π‘˜π‘₯1π‘Ÿπ‘˜subscriptπ‘„π‘˜π‘₯subscriptπ‘„π‘Ÿ1π‘₯superscriptπ‘Ÿ2subscriptπ‘„π‘Ÿπ‘˜1subscriptπ‘„π‘˜1π‘₯Q_{k}(x)=\frac{1}{r}kQ_{k}(x)+\frac{Q_{r-1}(x)}{r^{2}Q_{r}}(k+1)Q_{k+1}(x).

That is,

Qk​(x)=Crk​Qrβˆ’1r​(x)rrβˆ’k​Qrrβˆ’kβˆ’1.subscriptπ‘„π‘˜π‘₯superscriptsubscriptπΆπ‘Ÿπ‘˜superscriptsubscriptπ‘„π‘Ÿ1π‘Ÿπ‘₯superscriptπ‘Ÿπ‘Ÿπ‘˜superscriptsubscriptπ‘„π‘Ÿπ‘Ÿπ‘˜1Q_{k}(x)=C_{r}^{k}\frac{Q_{r-1}^{r}(x)}{r^{r-k}Q_{r}^{r-k-1}}.

Thus, we have that Qi​(x)=Cri​Qrβˆ’1rβˆ’i​(x)rrβˆ’i​Qrrβˆ’iβˆ’1​(x)subscript𝑄𝑖π‘₯superscriptsubscriptπΆπ‘Ÿπ‘–superscriptsubscriptπ‘„π‘Ÿ1π‘Ÿπ‘–π‘₯superscriptπ‘Ÿπ‘Ÿπ‘–superscriptsubscriptπ‘„π‘Ÿπ‘Ÿπ‘–1π‘₯Q_{i}(x)=C_{r}^{i}\frac{Q_{r-1}^{r-i}(x)}{r^{r-i}Q_{r}^{r-i-1}(x)} for 1≀i≀rβˆ’21π‘–π‘Ÿ21\leq i\leq r-2. Then equation (2.3)2.3(2.3) has the following form: Q0​(x)+c=Q1​(x)​Qrβˆ’1​(x)r2​Qrsubscript𝑄0π‘₯𝑐subscript𝑄1π‘₯subscriptπ‘„π‘Ÿ1π‘₯superscriptπ‘Ÿ2subscriptπ‘„π‘ŸQ_{0}(x)+c=Q_{1}(x)\frac{Q_{r-1}(x)}{r^{2}Q_{r}}. That is,

Q0​(x)=Qrβˆ’1r​(x)rr​Qrrβˆ’1βˆ’c.subscript𝑄0π‘₯superscriptsubscriptπ‘„π‘Ÿ1π‘Ÿπ‘₯superscriptπ‘Ÿπ‘Ÿsuperscriptsubscriptπ‘„π‘Ÿπ‘Ÿ1𝑐Q_{0}(x)=\frac{Q_{r-1}^{r}(x)}{r^{r}Q_{r}^{r-1}}-c.

Thus, we have Q​(x,y)=Qr​(yr+Qrβˆ’1​(x)Qr​yrβˆ’1+β‹―+Qrβˆ’1r​(x)rr​Qrr)βˆ’c=Qr​(y+Qrβˆ’1​(x)r​Qr)rβˆ’c𝑄π‘₯𝑦subscriptπ‘„π‘Ÿsuperscriptπ‘¦π‘Ÿsubscriptπ‘„π‘Ÿ1π‘₯subscriptπ‘„π‘Ÿsuperscriptπ‘¦π‘Ÿ1β‹―superscriptsubscriptπ‘„π‘Ÿ1π‘Ÿπ‘₯superscriptπ‘Ÿπ‘Ÿsuperscriptsubscriptπ‘„π‘Ÿπ‘Ÿπ‘subscriptπ‘„π‘Ÿsuperscript𝑦subscriptπ‘„π‘Ÿ1π‘₯π‘Ÿsubscriptπ‘„π‘Ÿπ‘Ÿπ‘Q(x,y)=Q_{r}(y^{r}+\frac{Q_{r-1}(x)}{Q_{r}}y^{r-1}+\cdots+\frac{Q_{r-1}^{r}(x)}{r^{r}Q_{r}^{r}})-c=Q_{r}(y+\frac{Q_{r-1}(x)}{rQ_{r}})^{r}-c. Therefore, we have QβˆˆπŠβ€‹[y+Qrβˆ’1​(x)r​Qr]π‘„πŠdelimited-[]𝑦subscriptπ‘„π‘Ÿ1π‘₯π‘Ÿsubscriptπ‘„π‘ŸQ\in{\bf K}[y+\frac{Q_{r-1}(x)}{rQ_{r}}]. Let a​(x)=Qrβˆ’1​(x)r​Qrπ‘Žπ‘₯subscriptπ‘„π‘Ÿ1π‘₯π‘Ÿsubscriptπ‘„π‘Ÿa(x)=\frac{Q_{r-1}(x)}{rQ_{r}}. Then the conclusion follows. ∎

Theorem 2.2.

Let H=(u​(x,y),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y),v(x,y,z),h(x,y)) be a polynomial map over πŠβ€‹[x,y,z]𝐊π‘₯𝑦𝑧{\bf K}[x,y,z]. Assume that H​(0)=0𝐻00H(0)=0 and the components of H𝐻H are linearly independent over 𝐊𝐊{\bf K}. If J​H𝐽𝐻JH is nilpotent and (degy⁑u,degy⁑h)≀3subscriptdegree𝑦𝑒subscriptdegreeπ‘¦β„Ž3(\deg_{y}u,\deg_{y}h)\leq 3, then u=g​(a​y+b​(x))π‘’π‘”π‘Žπ‘¦π‘π‘₯u=g(ay+b(x)), v=v1​zβˆ’aβˆ’1​b′​(x)​g​(a​y+b​(x))βˆ’v1​l2​x𝑣subscript𝑣1𝑧superscriptπ‘Ž1superscript𝑏′π‘₯π‘”π‘Žπ‘¦π‘π‘₯subscript𝑣1subscript𝑙2π‘₯v=v_{1}z-a^{-1}b^{\prime}(x)g(ay+b(x))-v_{1}l_{2}x, h=c0​u2+l2​uβ„Žsubscript𝑐0superscript𝑒2subscript𝑙2𝑒h=c_{0}u^{2}+l_{2}u, where b​(x)=v1​c0​a​x2+l1​x+l~2𝑏π‘₯subscript𝑣1subscript𝑐0π‘Žsuperscriptπ‘₯2subscript𝑙1π‘₯subscript~𝑙2b(x)=v_{1}c_{0}ax^{2}+l_{1}x+\tilde{l}_{2}; v1,c0,aβˆˆπŠβˆ—subscript𝑣1subscript𝑐0π‘Žsuperscript𝐊v_{1},c_{0},a\in{\bf K}^{*}; l1,l2,l~2∈𝐊subscript𝑙1subscript𝑙2subscript~𝑙2𝐊l_{1},l_{2},\tilde{l}_{2}\in{\bf K}, g​(t)βˆˆπŠβ€‹[t]π‘”π‘‘πŠdelimited-[]𝑑g(t)\in{\bf K}[t] and g​(0)=0𝑔00g(0)=0, degt⁑g​(t)β‰₯1subscriptdegree𝑑𝑔𝑑1\deg_{t}g(t)\geq 1.

Proof.

If (degy⁑u,degy⁑h)≀2subscriptdegree𝑦𝑒subscriptdegreeπ‘¦β„Ž2(\deg_{y}u,\deg_{y}h)\leq 2, then the conclusion follows from Theorem 3.5 in [13]. Let v=vd​zd+β‹―+v1​z+v0𝑣subscript𝑣𝑑superscript𝑧𝑑⋯subscript𝑣1𝑧subscript𝑣0v=v_{d}z^{d}+\cdots+v_{1}z+v_{0}. Then it follows from Lemma 3.2 in [13] that d=1𝑑1d=1 and v1βˆˆπŠβˆ—subscript𝑣1superscript𝐊v_{1}\in{\bf K}^{*}. Since J​H𝐽𝐻JH is nilpotent, so we have the following equations:

{ux+v0​y=0(2.4)ux​v0​yβˆ’v0​x​uyβˆ’v1​hy=0(2.5)v1​(ux​hyβˆ’uy​hx)=0(2.6)\left\{\begin{aligned} u_{x}+v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.4)\\ u_{x}v_{0y}-v_{0x}u_{y}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}(2.5)\\ v_{1}(u_{x}h_{y}-u_{y}h_{x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.6)\end{aligned}\right.

It follows from equation (2.4)2.4(2.4) that ux=βˆ’v0​ysubscript𝑒π‘₯subscript𝑣0𝑦u_{x}=-v_{0y}. Thus, there exists PβˆˆπŠβ€‹[x,y]π‘ƒπŠπ‘₯𝑦P\in{\bf K}[x,y] such that

u=βˆ’Py,v=Px(2.7)formulae-sequence𝑒subscript𝑃𝑦𝑣subscript𝑃π‘₯2.7u=-P_{y},~{}~{}v=P_{x}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.7)

It follows from equation (2.6)2.6(2.6) and Lemma 3.1 in [13] that there exists qβˆˆπŠβ€‹[x,y]π‘žπŠπ‘₯𝑦q\in{\bf K}[x,y] such that

u,hβˆˆπŠβ€‹[q](2.8)formulae-sequenceπ‘’β„ŽπŠdelimited-[]π‘ž2.8u,~{}h\in{\bf K}[q]~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.8)

Since degy⁑q|(degy⁑u,degy⁑h)conditionalsubscriptdegreeπ‘¦π‘žsubscriptdegree𝑦𝑒subscriptdegreeπ‘¦β„Ž\deg_{y}q|(\deg_{y}u,\deg_{y}h) and (degy⁑u,degy⁑h)=3subscriptdegree𝑦𝑒subscriptdegreeπ‘¦β„Ž3(\deg_{y}u,\deg_{y}h)=3, so we have degy⁑q=1subscriptdegreeπ‘¦π‘ž1\deg_{y}q=1 or 3.

If degy⁑q=1subscriptdegreeπ‘¦π‘ž1\deg_{y}q=1, then the conclusion follows from the proof of Theorem 2.8 in [12].

If degy⁑q=3subscriptdegreeπ‘¦π‘ž3\deg_{y}q=3, then degy⁑qy=2subscriptdegree𝑦subscriptπ‘žπ‘¦2\deg_{y}q_{y}=2. Let q​(x,y)=q3​(x)​y3+q2​(x)​y2+q1​(x)​y+q0​(x)π‘žπ‘₯𝑦subscriptπ‘ž3π‘₯superscript𝑦3subscriptπ‘ž2π‘₯superscript𝑦2subscriptπ‘ž1π‘₯𝑦subscriptπ‘ž0π‘₯q(x,y)=q_{3}(x)y^{3}+q_{2}(x)y^{2}+q_{1}(x)y+q_{0}(x).
Claim: q3​(x)βˆˆπŠβˆ—subscriptπ‘ž3π‘₯superscript𝐊q_{3}(x)\in{\bf K}^{*}.
Let P​(x,y)=ar​(x)​yr+arβˆ’1​(x)​yrβˆ’1+β‹―+a1​(x)​y+a0​(x)𝑃π‘₯𝑦subscriptπ‘Žπ‘Ÿπ‘₯superscriptπ‘¦π‘Ÿsubscriptπ‘Žπ‘Ÿ1π‘₯superscriptπ‘¦π‘Ÿ1β‹―subscriptπ‘Ž1π‘₯𝑦subscriptπ‘Ž0π‘₯P(x,y)=a_{r}(x)y^{r}+a_{r-1}(x)y^{r-1}+\cdots+a_{1}(x)y+a_{0}(x). It follows from Lemma 3.4 in [13] that ar​(x)βˆˆπŠβˆ—subscriptπ‘Žπ‘Ÿπ‘₯superscript𝐊a_{r}(x)\in{\bf K}^{*}. It follows from equations (2.7)2.7(2.7) and (2.8)2.8(2.8) that u​(x,y)=u​(q)=βˆ’Py=βˆ’(r​ar​(x)​yrβˆ’1+β‹―+a1​(x))𝑒π‘₯π‘¦π‘’π‘žsubscriptπ‘ƒπ‘¦π‘Ÿsubscriptπ‘Žπ‘Ÿπ‘₯superscriptπ‘¦π‘Ÿ1β‹―subscriptπ‘Ž1π‘₯u(x,y)=u(q)=-P_{y}=-(ra_{r}(x)y^{r-1}+\cdots+a_{1}(x)). Thus, we have q3​(x)βˆˆπŠβˆ—subscriptπ‘ž3π‘₯superscript𝐊q_{3}(x)\in{\bf K}^{*}. Substituting equations (2.7)2.7(2.7) and (2.8)2.8(2.8) to equation (2.5)2.5(2.5), we have the following equation:

qy​[v1​h′​(q)+v0​x​u′​(q)]=βˆ’(u′​(q)​qx)2(2.9)subscriptπ‘žπ‘¦delimited-[]subscript𝑣1superscriptβ„Žβ€²π‘žsubscript𝑣0π‘₯superscriptπ‘’β€²π‘žsuperscriptsuperscriptπ‘’β€²π‘žsubscriptπ‘žπ‘₯22.9q_{y}[v_{1}h^{\prime}(q)+v_{0x}u^{\prime}(q)]=-(u^{\prime}(q)q_{x})^{2}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.9)

(1) If qysubscriptπ‘žπ‘¦q_{y} is irreducible, then we have qy|qxconditionalsubscriptπ‘žπ‘¦subscriptπ‘žπ‘₯q_{y}|q_{x} or qy|u′​(q)conditionalsubscriptπ‘žπ‘¦superscriptπ‘’β€²π‘žq_{y}|u^{\prime}(q). Since u′​(q)superscriptπ‘’β€²π‘žu^{\prime}(q) is a polynomial of qπ‘žq, so we have u′​(q)=c0​(q+c1)​(q+c2)​⋯​(q+ck)superscriptπ‘’β€²π‘žsubscript𝑐0π‘žsubscript𝑐1π‘žsubscript𝑐2β‹―π‘žsubscriptπ‘π‘˜u^{\prime}(q)=c_{0}(q+c_{1})(q+c_{2})\cdots(q+c_{k}) with c0βˆˆπŠβˆ—subscript𝑐0superscript𝐊c_{0}\in{\bf K}^{*} and ci∈𝐊subscriptπ‘π‘–πŠc_{i}\in{\bf K}, 1≀i≀k1π‘–π‘˜1\leq i\leq k. Since qysubscriptπ‘žπ‘¦q_{y} is irreducible, so there exists i0∈{1,2,…,k}subscript𝑖012β€¦π‘˜i_{0}\in\{1,2,\ldots,k\} such that qy|q+ci0conditionalsubscriptπ‘žπ‘¦π‘žsubscript𝑐subscript𝑖0q_{y}|q+c_{i_{0}} in the case that qy|u′​(q)conditionalsubscriptπ‘žπ‘¦superscriptπ‘’β€²π‘žq_{y}|u^{\prime}(q). It follows from Lemma 2.1 that qπ‘žq is a polynomial of y+a​(x)π‘¦π‘Žπ‘₯y+a(x) for some a​(x)βˆˆπŠβ€‹[x]π‘Žπ‘₯𝐊delimited-[]π‘₯a(x)\in{\bf K}[x]. That is, u,hβˆˆπŠβ€‹[y+a​(x)]π‘’β„ŽπŠdelimited-[]π‘¦π‘Žπ‘₯u,h\in{\bf K}[y+a(x)]. Then the conclusion follows from the proof of Theorem 2.8 in [12].

(2) If qysubscriptπ‘žπ‘¦q_{y} is reducible, then

qy=3​q3​y2+2​q2​(x)​y+q1​(x)=3​q3​(y+b​(x))​(y+e​(x))(2.10)formulae-sequencesubscriptπ‘žπ‘¦3subscriptπ‘ž3superscript𝑦22subscriptπ‘ž2π‘₯𝑦subscriptπ‘ž1π‘₯3subscriptπ‘ž3𝑦𝑏π‘₯𝑦𝑒π‘₯2.10q_{y}=3q_{3}y^{2}+2q_{2}(x)y+q_{1}(x)=3q_{3}(y+b(x))(y+e(x))~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.10)

for some b​(x),e​(x)βˆˆπŠβ€‹[x]𝑏π‘₯𝑒π‘₯𝐊delimited-[]π‘₯b(x),~{}e(x)\in{\bf K}[x]. Clearly, y+b​(x)𝑦𝑏π‘₯y+b(x), y+e​(x)𝑦𝑒π‘₯y+e(x) are irreducible. It follows from equation (2.9)2.9(2.9) that we have the following cases:
Case I qy|qxconditionalsubscriptπ‘žπ‘¦subscriptπ‘žπ‘₯q_{y}|q_{x}
Case II qy|u′​(q)conditionalsubscriptπ‘žπ‘¦superscriptπ‘’β€²π‘žq_{y}|u^{\prime}(q)
Case III (y+b​(x))|qxconditional𝑦𝑏π‘₯subscriptπ‘žπ‘₯(y+b(x))|q_{x} and (y+e​(x))|u′​(q)conditional𝑦𝑒π‘₯superscriptπ‘’β€²π‘ž(y+e(x))|u^{\prime}(q)
Case IV (y+b​(x))|qxconditional𝑦𝑏π‘₯subscriptπ‘žπ‘₯(y+b(x))|q_{x} in the case b​(x)=e​(x)𝑏π‘₯𝑒π‘₯b(x)=e(x).
Case V (y+b​(x))|u′​(q)conditional𝑦𝑏π‘₯superscriptπ‘’β€²π‘ž(y+b(x))|u^{\prime}(q) in the case b​(x)=e​(x)𝑏π‘₯𝑒π‘₯b(x)=e(x).

Case I If qy|qxconditionalsubscriptπ‘žπ‘¦subscriptπ‘žπ‘₯q_{y}|q_{x}, then the conclusion follows from Lemma 2.1.

Case II If qy|u′​(q)conditionalsubscriptπ‘žπ‘¦superscriptπ‘’β€²π‘žq_{y}|u^{\prime}(q), then (y+b​(x))​(y+e​(x))|u′​(q)conditional𝑦𝑏π‘₯𝑦𝑒π‘₯superscriptπ‘’β€²π‘ž(y+b(x))(y+e(x))|u^{\prime}(q). Since u′​(q)superscriptπ‘’β€²π‘žu^{\prime}(q) is a polynomial of qπ‘žq, so u′​(q)=c0​(q+c1)​(q+c2)​⋯​(q+ck)superscriptπ‘’β€²π‘žsubscript𝑐0π‘žsubscript𝑐1π‘žsubscript𝑐2β‹―π‘žsubscriptπ‘π‘˜u^{\prime}(q)=c_{0}(q+c_{1})(q+c_{2})\cdots(q+c_{k}) for c0βˆˆπŠβˆ—subscript𝑐0superscript𝐊c_{0}\in{\bf K}^{*} and ci∈𝐊subscriptπ‘π‘–πŠc_{i}\in{\bf K}, 1≀i≀k1π‘–π‘˜1\leq i\leq k. Thus, there exist i0,j0∈{1,2,…,k}subscript𝑖0subscript𝑗012β€¦π‘˜i_{0},~{}j_{0}\in\{1,2,\ldots,k\} such that (y+b​(x))|q+ci0conditional𝑦𝑏π‘₯π‘žsubscript𝑐subscript𝑖0(y+b(x))|q+c_{i_{0}} and (y+e​(x))|q+cj0conditional𝑦𝑒π‘₯π‘žsubscript𝑐subscript𝑗0(y+e(x))|q+c_{j_{0}}. That is,

q3​y3+q2​(x)​y2+q1​(x)​y+q0​(x)+ci0=(y+b​(x))​(q3​y2+d1​(x)​y+d0​(x))(2.11)subscriptπ‘ž3superscript𝑦3subscriptπ‘ž2π‘₯superscript𝑦2subscriptπ‘ž1π‘₯𝑦subscriptπ‘ž0π‘₯subscript𝑐subscript𝑖0𝑦𝑏π‘₯subscriptπ‘ž3superscript𝑦2subscript𝑑1π‘₯𝑦subscript𝑑0π‘₯2.11q_{3}y^{3}+q_{2}(x)y^{2}+q_{1}(x)y+q_{0}(x)+c_{i_{0}}=(y+b(x))(q_{3}y^{2}+d_{1}(x)y+d_{0}(x))~{}~{}~{}~{}~{}(2.11)

and

q3​y3+q2​(x)​y2+q1​(x)​y+q0​(x)+cj0=(y+e​(x))​(q3​y2+f1​(x)​y+f0​(x))(2.12)subscriptπ‘ž3superscript𝑦3subscriptπ‘ž2π‘₯superscript𝑦2subscriptπ‘ž1π‘₯𝑦subscriptπ‘ž0π‘₯subscript𝑐subscript𝑗0𝑦𝑒π‘₯subscriptπ‘ž3superscript𝑦2subscript𝑓1π‘₯𝑦subscript𝑓0π‘₯2.12q_{3}y^{3}+q_{2}(x)y^{2}+q_{1}(x)y+q_{0}(x)+c_{j_{0}}=(y+e(x))(q_{3}y^{2}+f_{1}(x)y+f_{0}(x))~{}~{}~{}~{}~{}(2.12)

We always view that the polynomials are in πŠβ€‹[x]​[y]𝐊delimited-[]π‘₯delimited-[]𝑦{\bf K}[x][y] with coefficients in πŠβ€‹[x]𝐊delimited-[]π‘₯{\bf K}[x] in the following arguments. Comparing the coefficients of y2superscript𝑦2y^{2}, y𝑦y of equations (2.11)2.11(2.11) and (2.12)2.12(2.12), we have the following equations:

q2​(x)=d1​(x)+q3​b​(x)=f1​(x)+q3​e​(x)(2.13)formulae-sequencesubscriptπ‘ž2π‘₯subscript𝑑1π‘₯subscriptπ‘ž3𝑏π‘₯subscript𝑓1π‘₯subscriptπ‘ž3𝑒π‘₯2.13q_{2}(x)=d_{1}(x)+q_{3}b(x)=f_{1}(x)+q_{3}e(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.13)

and

q1​(x)=d0​(x)+d1​(x)​b​(x)=f0​(x)+e​(x)​f1​(x)(2.14)formulae-sequencesubscriptπ‘ž1π‘₯subscript𝑑0π‘₯subscript𝑑1π‘₯𝑏π‘₯subscript𝑓0π‘₯𝑒π‘₯subscript𝑓1π‘₯2.14q_{1}(x)=d_{0}(x)+d_{1}(x)b(x)=f_{0}(x)+e(x)f_{1}(x)~{}~{}~{}~{}~{}(2.14)

It follows from equations (2.13)2.13(2.13) and (2.14)2.14(2.14) that

d0​(x)=q1​(x)βˆ’b​(x)​q2​(x)+q3​b2​(x)(2.15)subscript𝑑0π‘₯subscriptπ‘ž1π‘₯𝑏π‘₯subscriptπ‘ž2π‘₯subscriptπ‘ž3superscript𝑏2π‘₯2.15d_{0}(x)=q_{1}(x)-b(x)q_{2}(x)+q_{3}b^{2}(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.15)

and

f0​(x)=q1​(x)βˆ’e​(x)​q2​(x)+q3​e2​(x)(2.16)subscript𝑓0π‘₯subscriptπ‘ž1π‘₯𝑒π‘₯subscriptπ‘ž2π‘₯subscriptπ‘ž3superscript𝑒2π‘₯2.16f_{0}(x)=q_{1}(x)-e(x)q_{2}(x)+q_{3}e^{2}(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.16)

Then equations (2.11)2.11(2.11) and (2.12)2.12(2.12) have the following forms:

q0​(x)+ci0=b​(x)​d0​(x)=b​(x)​(q1​(x)βˆ’b​(x)​q2​(x)+q3​b2​(x))(2.17)formulae-sequencesubscriptπ‘ž0π‘₯subscript𝑐subscript𝑖0𝑏π‘₯subscript𝑑0π‘₯𝑏π‘₯subscriptπ‘ž1π‘₯𝑏π‘₯subscriptπ‘ž2π‘₯subscriptπ‘ž3superscript𝑏2π‘₯2.17q_{0}(x)+c_{i_{0}}=b(x)d_{0}(x)=b(x)(q_{1}(x)-b(x)q_{2}(x)+q_{3}b^{2}(x))~{}~{}~{}~{}~{}~{}~{}~{}(2.17)

and

q0​(x)+cj0=e​(x)​f0​(x)=e​(x)​(q1​(x)βˆ’e​(x)​q2​(x)+q3​e2​(x))(2.18)formulae-sequencesubscriptπ‘ž0π‘₯subscript𝑐subscript𝑗0𝑒π‘₯subscript𝑓0π‘₯𝑒π‘₯subscriptπ‘ž1π‘₯𝑒π‘₯subscriptπ‘ž2π‘₯subscriptπ‘ž3superscript𝑒2π‘₯2.18q_{0}(x)+c_{j_{0}}=e(x)f_{0}(x)=e(x)(q_{1}(x)-e(x)q_{2}(x)+q_{3}e^{2}(x))~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.18)

Then equation (2.17)βˆ’(2.18)2.172.18(2.17)-(2.18) has the following form:

ci0βˆ’cj0=(b​(x)βˆ’e​(x))​[q1​(x)βˆ’q2​(x)​(b​(x)+e​(x))+q3​(b2​(x)+b​(x)​e​(x)+e2​(x))].subscript𝑐subscript𝑖0subscript𝑐subscript𝑗0𝑏π‘₯𝑒π‘₯delimited-[]subscriptπ‘ž1π‘₯subscriptπ‘ž2π‘₯𝑏π‘₯𝑒π‘₯subscriptπ‘ž3superscript𝑏2π‘₯𝑏π‘₯𝑒π‘₯superscript𝑒2π‘₯c_{i_{0}}-c_{j_{0}}=(b(x)-e(x))[q_{1}(x)-q_{2}(x)(b(x)+e(x))+q_{3}(b^{2}(x)+b(x)e(x)+e^{2}(x))].

Thus, we have b​(x)βˆ’e​(x)βˆˆπŠπ‘π‘₯𝑒π‘₯𝐊b(x)-e(x)\in{\bf K}. That is, e​(x)=b​(x)+c𝑒π‘₯𝑏π‘₯𝑐e(x)=b(x)+c for some cβˆˆπŠπ‘πŠc\in{\bf K}. Therefore, we have qy=3​q3​(y+b​(x))​(y+b​(x)+c)=3​q3​(y+b​(x))2+3​q3​c​(y+b​(x))subscriptπ‘žπ‘¦3subscriptπ‘ž3𝑦𝑏π‘₯𝑦𝑏π‘₯𝑐3subscriptπ‘ž3superscript𝑦𝑏π‘₯23subscriptπ‘ž3𝑐𝑦𝑏π‘₯q_{y}=3q_{3}(y+b(x))(y+b(x)+c)=3q_{3}(y+b(x))^{2}+3q_{3}c(y+b(x)). Then q​(x,y)=q3​(y+b​(x))3+32​q3​c​(y+b​(x))2+f​(x)π‘žπ‘₯𝑦subscriptπ‘ž3superscript𝑦𝑏π‘₯332subscriptπ‘ž3𝑐superscript𝑦𝑏π‘₯2𝑓π‘₯q(x,y)=q_{3}(y+b(x))^{3}+\frac{3}{2}q_{3}c(y+b(x))^{2}+f(x) for some f​(x)βˆˆπŠβ€‹[x]𝑓π‘₯𝐊delimited-[]π‘₯f(x)\in{\bf K}[x]. Since (y+b​(x))|(q+ci0)conditional𝑦𝑏π‘₯π‘žsubscript𝑐subscript𝑖0(y+b(x))|(q+c_{i_{0}}), so we have (y+b​(x))|(f​(x)+ci0)conditional𝑦𝑏π‘₯𝑓π‘₯subscript𝑐subscript𝑖0(y+b(x))|(f(x)+c_{i_{0}}). That is, f​(x)=βˆ’ci0𝑓π‘₯subscript𝑐subscript𝑖0f(x)=-c_{i_{0}}. Thus, qπ‘žq is a polynomial of y+b​(x)𝑦𝑏π‘₯y+b(x). Therefore, u,hβˆˆπŠβ€‹[y+b​(x)]π‘’β„ŽπŠdelimited-[]𝑦𝑏π‘₯u,h\in{\bf K}[y+b(x)]. Then the conclusion follows from the proof of Theorem 2.8 in [12]

Case III Since u′​(q)=c0​(q+c1)​(q+c2)​⋯​(q+ck)superscriptπ‘’β€²π‘žsubscript𝑐0π‘žsubscript𝑐1π‘žsubscript𝑐2β‹―π‘žsubscriptπ‘π‘˜u^{\prime}(q)=c_{0}(q+c_{1})(q+c_{2})\cdots(q+c_{k}) for c0βˆˆπŠβˆ—subscript𝑐0superscript𝐊c_{0}\in{\bf K}^{*} and ci∈𝐊subscriptπ‘π‘–πŠc_{i}\in{\bf K}, 1≀i≀k1π‘–π‘˜1\leq i\leq k. Clearly, y+e​(x)𝑦𝑒π‘₯y+e(x) is irreducible, so there exists i0∈{1,2,…,k}subscript𝑖012β€¦π‘˜i_{0}\in\{1,2,\ldots,k\} such that (y+e​(x))|(q+ci0)conditional𝑦𝑒π‘₯π‘žsubscript𝑐subscript𝑖0(y+e(x))|(q+c_{i_{0}}). That is,

q2′​(x)​y2+q1′​(x)​y+q0′​(x)=(y+b​(x))​(q2′​(x)​y+d​(x))(2.19)subscriptsuperscriptπ‘žβ€²2π‘₯superscript𝑦2superscriptsubscriptπ‘ž1β€²π‘₯𝑦subscriptsuperscriptπ‘žβ€²0π‘₯𝑦𝑏π‘₯subscriptsuperscriptπ‘žβ€²2π‘₯𝑦𝑑π‘₯2.19q^{\prime}_{2}(x)y^{2}+q_{1}^{\prime}(x)y+q^{\prime}_{0}(x)=(y+b(x))(q^{\prime}_{2}(x)y+d(x))~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.19)

and

q3​y3+q2​(x)​y2+q1​(x)​y+q0​(x)+ci0=(y+e​(x))​(q3​y2+w1​(x)​y+w0​(x))(2.20)subscriptπ‘ž3superscript𝑦3subscriptπ‘ž2π‘₯superscript𝑦2subscriptπ‘ž1π‘₯𝑦subscriptπ‘ž0π‘₯subscript𝑐subscript𝑖0𝑦𝑒π‘₯subscriptπ‘ž3superscript𝑦2subscript𝑀1π‘₯𝑦subscript𝑀0π‘₯2.20q_{3}y^{3}+q_{2}(x)y^{2}+q_{1}(x)y+q_{0}(x)+c_{i_{0}}=(y+e(x))(q_{3}y^{2}+w_{1}(x)y+w_{0}(x))~{}~{}~{}~{}~{}~{}(2.20)

where d​(x),w1​(x),w0​(x)βˆˆπŠβ€‹[x]𝑑π‘₯subscript𝑀1π‘₯subscript𝑀0π‘₯𝐊delimited-[]π‘₯d(x),w_{1}(x),w_{0}(x)\in{\bf K}[x]. Comparing the coefficients of y𝑦y, y0superscript𝑦0y^{0} of equation (2.19)2.19(2.19), we have the following equations:

q1′​(x)βˆ’b​(x)​q2′​(x)=d​(x)subscriptsuperscriptπ‘žβ€²1π‘₯𝑏π‘₯subscriptsuperscriptπ‘žβ€²2π‘₯𝑑π‘₯q^{\prime}_{1}(x)-b(x)q^{\prime}_{2}(x)=d(x)

and

q0′​(x)=b​(x)​d​(x)=b​(x)​(q1′​(x)βˆ’b​(x)​q2′​(x))(2.21)formulae-sequencesubscriptsuperscriptπ‘žβ€²0π‘₯𝑏π‘₯𝑑π‘₯𝑏π‘₯subscriptsuperscriptπ‘žβ€²1π‘₯𝑏π‘₯subscriptsuperscriptπ‘žβ€²2π‘₯2.21q^{\prime}_{0}(x)=b(x)d(x)=b(x)(q^{\prime}_{1}(x)-b(x)q^{\prime}_{2}(x))~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.21)

Comparing the coefficients of y𝑦y, y0superscript𝑦0y^{0} of equation (2.10)2.10(2.10), we have the following equations:

2​q2​(x)3​q3=b​(x)+e​(x)(2.22)2subscriptπ‘ž2π‘₯3subscriptπ‘ž3𝑏π‘₯𝑒π‘₯2.22\frac{2q_{2}(x)}{3q_{3}}=b(x)+e(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.22)

and

q1​(x)3​q3=b​(x)​e​(x)(2.23)subscriptπ‘ž1π‘₯3subscriptπ‘ž3𝑏π‘₯𝑒π‘₯2.23\frac{q_{1}(x)}{3q_{3}}=b(x)e(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.23)

Substituting equations (2.22)2.22(2.22) and (2.23)2.23(2.23) to equation (2.21)2.21(2.21), we have the following equation:

q0′​(x)=q3​(3​b​(x)​b′​(x)​e​(x)+32​b2​(x)​e′​(x)βˆ’32​b2​(x)​b′​(x))subscriptsuperscriptπ‘žβ€²0π‘₯subscriptπ‘ž33𝑏π‘₯superscript𝑏′π‘₯𝑒π‘₯32superscript𝑏2π‘₯superscript𝑒′π‘₯32superscript𝑏2π‘₯superscript𝑏′π‘₯q^{\prime}_{0}(x)=q_{3}(3b(x)b^{\prime}(x)e(x)+\frac{3}{2}b^{2}(x)e^{\prime}(x)-\frac{3}{2}b^{2}(x)b^{\prime}(x))

Integrating the two sides of the above equation with respect to xπ‘₯x, we have the following equation:

q0​(x)=q3​(32​b2​(x)​e​(x)βˆ’12​b3​(x))+cΒ―(2.24)subscriptπ‘ž0π‘₯subscriptπ‘ž332superscript𝑏2π‘₯𝑒π‘₯12superscript𝑏3π‘₯¯𝑐2.24q_{0}(x)=q_{3}(\frac{3}{2}b^{2}(x)e(x)-\frac{1}{2}b^{3}(x))+\bar{c}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.24)

for some cΒ―βˆˆπŠΒ―π‘πŠ\bar{c}\in{\bf K}. Comparing the coefficients of y2superscript𝑦2y^{2}, y𝑦y and y0superscript𝑦0y^{0} of equation (2.20)2.20(2.20), we have the following equations:

{q2​(x)=w1​(x)+q3​e​(x)(2.25)q1​(x)=w0​(x)+w1​(x)​e​(x)(2.26)q0​(x)+ci0=e​(x)​w0​(x)(2.27)\left\{\begin{aligned} q_{2}(x)=w_{1}(x)+q_{3}e(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.25)\\ q_{1}(x)=w_{0}(x)+w_{1}(x)e(x)~{}~{}~{}~{}~{}~{}~{}(2.26)\\ q_{0}(x)+c_{i_{0}}=e(x)w_{0}(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.27)\\ \end{aligned}\right.

It follows from equations (2.25)2.25(2.25) and (2.26)2.26(2.26) that

w0​(x)=q1​(x)βˆ’e​(x)​q2​(x)+q3​e2​(x)(2.28)subscript𝑀0π‘₯subscriptπ‘ž1π‘₯𝑒π‘₯subscriptπ‘ž2π‘₯subscriptπ‘ž3superscript𝑒2π‘₯2.28w_{0}(x)=q_{1}(x)-e(x)q_{2}(x)+q_{3}e^{2}(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.28)

Substituting equations (2.28)2.28(2.28), (2.22)2.22(2.22) and (2.23)2.23(2.23) to equation (2.27)2.27(2.27), we have the following equation:

q0​(x)=q3​(32​b​(x)​e2​(x)βˆ’12​e3​(x))βˆ’ci0(2.29)subscriptπ‘ž0π‘₯subscriptπ‘ž332𝑏π‘₯superscript𝑒2π‘₯12superscript𝑒3π‘₯subscript𝑐subscript𝑖02.29q_{0}(x)=q_{3}(\frac{3}{2}b(x)e^{2}(x)-\frac{1}{2}e^{3}(x))-c_{i_{0}}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.29)

Comparing equation (2.29)2.29(2.29) with equation (2.24)2.24(2.24), we have 12​q3​[b3​(x)βˆ’3​b2​(x)​e​(x)+3​b​(x)​e2​(x)βˆ’e3​(x)]=cΒ―+ci012subscriptπ‘ž3delimited-[]superscript𝑏3π‘₯3superscript𝑏2π‘₯𝑒π‘₯3𝑏π‘₯superscript𝑒2π‘₯superscript𝑒3π‘₯¯𝑐subscript𝑐subscript𝑖0\frac{1}{2}q_{3}[b^{3}(x)-3b^{2}(x)e(x)+3b(x)e^{2}(x)-e^{3}(x)]=\bar{c}+c_{i_{0}}. That is,

12​q3​(b​(x)βˆ’e​(x))3=cΒ―+ci0.12subscriptπ‘ž3superscript𝑏π‘₯𝑒π‘₯3¯𝑐subscript𝑐subscript𝑖0\frac{1}{2}q_{3}(b(x)-e(x))^{3}=\bar{c}+c_{i_{0}}.

Thus, we have b​(x)βˆ’e​(x)βˆˆπŠπ‘π‘₯𝑒π‘₯𝐊b(x)-e(x)\in{\bf K}. That is, b​(x)=e​(x)+c¯¯𝑏π‘₯𝑒π‘₯¯¯𝑐b(x)=e(x)+\bar{\bar{c}} for some cΒ―Β―βˆˆπŠΒ―Β―π‘πŠ\bar{\bar{c}}\in{\bf K}. It follows from equation (2.10)2.10(2.10) that qy=3​q3​(y+e​(x))2+3​q3​c¯¯​(y+e​(x))subscriptπ‘žπ‘¦3subscriptπ‘ž3superscript𝑦𝑒π‘₯23subscriptπ‘ž3¯¯𝑐𝑦𝑒π‘₯q_{y}=3q_{3}(y+e(x))^{2}+3q_{3}\bar{\bar{c}}(y+e(x)). Then we have q=q3​(y+e​(x))3+32​q3​c¯¯​(y+e​(x))2+f¯​(x)π‘žsubscriptπ‘ž3superscript𝑦𝑒π‘₯332subscriptπ‘ž3¯¯𝑐superscript𝑦𝑒π‘₯2¯𝑓π‘₯q=q_{3}(y+e(x))^{3}+\frac{3}{2}q_{3}\bar{\bar{c}}(y+e(x))^{2}+\bar{f}(x) for some f¯​(x)βˆˆπŠβ€‹[x]¯𝑓π‘₯𝐊delimited-[]π‘₯\bar{f}(x)\in{\bf K}[x]. Since (y+e​(x))|(q+ci0)conditional𝑦𝑒π‘₯π‘žsubscript𝑐subscript𝑖0(y+e(x))|(q+c_{i_{0}}), so we have (y+e​(x))|(f¯​(x)+ci0)conditional𝑦𝑒π‘₯¯𝑓π‘₯subscript𝑐subscript𝑖0(y+e(x))|(\bar{f}(x)+c_{i_{0}}). Thus, we have f¯​(x)=βˆ’ci0¯𝑓π‘₯subscript𝑐subscript𝑖0\bar{f}(x)=-c_{i_{0}}. Therefore, qπ‘žq is a polynomial of y+e​(x)𝑦𝑒π‘₯y+e(x). That is, u,h,qβˆˆπŠβ€‹[y+e​(x)]π‘’β„Žπ‘žπŠdelimited-[]𝑦𝑒π‘₯u,h,q\in{\bf K}[y+e(x)]. Then the conclusion follows from the proof of Theorem 2.8 in [12].

Case IV, V If b​(x)=e​(x)𝑏π‘₯𝑒π‘₯b(x)=e(x), then it follows from equation (2.10)2.10(2.10) that qy=3​q3​(y+b​(x))2subscriptπ‘žπ‘¦3subscriptπ‘ž3superscript𝑦𝑏π‘₯2q_{y}=3q_{3}(y+b(x))^{2}. Thus, we have q=q3​(y+b​(x))3+f¯¯​(x)π‘žsubscriptπ‘ž3superscript𝑦𝑏π‘₯3¯¯𝑓π‘₯q=q_{3}(y+b(x))^{3}+\bar{\bar{f}}(x) for some f¯¯​(x)βˆˆπŠβ€‹[x]¯¯𝑓π‘₯𝐊delimited-[]π‘₯\bar{\bar{f}}(x)\in{\bf K}[x].

If (y+b​(x))|qxconditional𝑦𝑏π‘₯subscriptπ‘žπ‘₯(y+b(x))|q_{x}, then (y+b​(x))|(3​q3​b′​(x)​(y+b​(x))2+f¯¯′​(x))conditional𝑦𝑏π‘₯3subscriptπ‘ž3superscript𝑏′π‘₯superscript𝑦𝑏π‘₯2superscript¯¯𝑓′π‘₯(y+b(x))|(3q_{3}b^{\prime}(x)(y+b(x))^{2}+\bar{\bar{f}}^{\prime}(x)). Thus, we have f¯¯′​(x)=0superscript¯¯𝑓′π‘₯0\bar{\bar{f}}^{\prime}(x)=0. That is, f¯¯​(x)βˆˆπŠΒ―Β―π‘“π‘₯𝐊\bar{\bar{f}}(x)\in{\bf K}. Therefore, we have qβˆˆπŠβ€‹[y+b​(x)]π‘žπŠdelimited-[]𝑦𝑏π‘₯q\in{\bf K}[y+b(x)].

If (y+b​(x))|u′​(q)conditional𝑦𝑏π‘₯superscriptπ‘’β€²π‘ž(y+b(x))|u^{\prime}(q), then it follows from the arguments of Case II that (y+b​(x))|q+ci0conditional𝑦𝑏π‘₯π‘žsubscript𝑐subscript𝑖0(y+b(x))|q+c_{i_{0}} for some ci0∈𝐊subscript𝑐subscript𝑖0𝐊c_{i_{0}}\in{\bf K}. That is, (y+b​(x))|(q3​(y+b​(x))3+f¯¯​(x)+ci0)conditional𝑦𝑏π‘₯subscriptπ‘ž3superscript𝑦𝑏π‘₯3¯¯𝑓π‘₯subscript𝑐subscript𝑖0(y+b(x))|(q_{3}(y+b(x))^{3}+\bar{\bar{f}}(x)+c_{i_{0}}). Thus, we have f¯¯​(x)=βˆ’ci0βˆˆπŠΒ―Β―π‘“π‘₯subscript𝑐subscript𝑖0𝐊\bar{\bar{f}}(x)=-c_{i_{0}}\in{\bf K}. Therefore, qβˆˆπŠβ€‹[y+b​(x)]π‘žπŠdelimited-[]𝑦𝑏π‘₯q\in{\bf K}[y+b(x)].
Thus, u,hπ‘’β„Žu,~{}h are polynomials of y+b​(x)𝑦𝑏π‘₯y+b(x) in the two cases. Then the conclusion follows from the proof of Theorem 2.8 in [12]. ∎

Remark 2.3.

We can replace the condition that (degy⁑u​(x,y),m)=1subscriptdegree𝑦𝑒π‘₯π‘¦π‘š1(\deg_{y}u(x,y),m)=1 by the condition (degy⁑u​(x,y),m)≀3subscriptdegree𝑦𝑒π‘₯π‘¦π‘š3(\deg_{y}u(x,y),m)\leq 3 in Theorem 2.10 and replace the condition that (m,n)=1π‘šπ‘›1(m,n)=1 by the condition (m,n)≀3π‘šπ‘›3(m,n)\leq 3 in Theorem 3.2 and Theorem 3.4 in [12].

Corollary 2.4.

Let H=(u​(x,y),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y),v(x,y,z),h(x,y)) be a polynomial map over πŠβ€‹[x,y,z]𝐊π‘₯𝑦𝑧{\bf K}[x,y,z]. Assume that H​(0)=0𝐻00H(0)=0 and the components of H𝐻H are linearly independent over 𝐊𝐊{\bf K}. If J​H𝐽𝐻JH is nilpotent and degy⁑u≀7subscriptdegree𝑦𝑒7\deg_{y}u\leq 7 or degy⁑u=9subscriptdegree𝑦𝑒9\deg_{y}u=9 or degy⁑h≀7subscriptdegreeπ‘¦β„Ž7\deg_{y}h\leq 7 or degy⁑h=9subscriptdegreeπ‘¦β„Ž9\deg_{y}h=9, then H𝐻H has the form of Theorem 2.2.

Proof.

Let v=vd​zd+β‹―+v1​z+v0𝑣subscript𝑣𝑑superscript𝑧𝑑⋯subscript𝑣1𝑧subscript𝑣0v=v_{d}z^{d}+\cdots+v_{1}z+v_{0}. Then it follows from Lemma 3.2 in [13] that d=1𝑑1d=1 and v1βˆˆπŠβˆ—subscript𝑣1superscript𝐊v_{1}\in{\bf K}^{*}. Since J​H𝐽𝐻JH is nilpotent, so we have the following equations:

{ux+v0​y=0(2.4)ux​v0​yβˆ’v0​x​uyβˆ’v1​hy=0(2.5)v1​(ux​hyβˆ’uy​hx)=0(2.6)\left\{\begin{aligned} u_{x}+v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.4)\\ u_{x}v_{0y}-v_{0x}u_{y}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}(2.5)\\ v_{1}(u_{x}h_{y}-u_{y}h_{x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.6)\end{aligned}\right.

It follows from equation (2.6)2.6(2.6) and Lemma 3.1 in [13] that there exists qβˆˆπŠβ€‹[x,y]π‘žπŠπ‘₯𝑦q\in{\bf K}[x,y] such that

u,hβˆˆπŠβ€‹[q](2.8)formulae-sequenceπ‘’β„ŽπŠdelimited-[]π‘ž2.8u,~{}h\in{\bf K}[q]~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.8)

Thus, we have degy⁑q|(degy⁑u,degy⁑h)conditionalsubscriptdegreeπ‘¦π‘žsubscriptdegree𝑦𝑒subscriptdegreeπ‘¦β„Ž\deg_{y}q|(\deg_{y}u,\deg_{y}h).

If degy⁑u≀4subscriptdegree𝑦𝑒4\deg_{y}u\leq 4 or degy⁑h≀4subscriptdegreeπ‘¦β„Ž4\deg_{y}h\leq 4, then the conclusion follows from Corollary 3.9 in [13].

If degy⁑u=5subscriptdegree𝑦𝑒5\deg_{y}u=5 or degy⁑u=7subscriptdegree𝑦𝑒7\deg_{y}u=7 or degy⁑h=5subscriptdegreeπ‘¦β„Ž5\deg_{y}h=5 or degy⁑h=7subscriptdegreeπ‘¦β„Ž7\deg_{y}h=7, then the conclusion follows from Theorem 3.3 in [13].

If degy⁑u=6subscriptdegree𝑦𝑒6\deg_{y}u=6 or 9 or degy⁑h=6subscriptdegreeπ‘¦β„Ž6\deg_{y}h=6 or 9, then degy⁑q=1subscriptdegreeπ‘¦π‘ž1\deg_{y}q=1 or 2 or 3 or 6 or 9.

Case I If degy⁑q=1subscriptdegreeπ‘¦π‘ž1\deg_{y}q=1 or 2 or 3, then the conclusion follows from the proof Theorem 2.2.

Case II If degy⁑q=6subscriptdegreeπ‘¦π‘ž6\deg_{y}q=6, then degy⁑u=6subscriptdegree𝑦𝑒6\deg_{y}u=6 or degy⁑h=6subscriptdegreeπ‘¦β„Ž6\deg_{y}h=6.

(1) If degy⁑q=degy⁑usubscriptdegreeπ‘¦π‘žsubscriptdegree𝑦𝑒\deg_{y}q=\deg_{y}u, then it follows from equation (2.8)2.8(2.8) that u​(x,y)=u​(q)=λ​q+Ξ»0𝑒π‘₯π‘¦π‘’π‘žπœ†π‘žsubscriptπœ†0u(x,y)=u(q)=\lambda q+\lambda_{0} with Ξ»βˆˆπŠβˆ—πœ†superscript𝐊\lambda\in{\bf K}^{*}, Ξ»0∈𝐊subscriptπœ†0𝐊\lambda_{0}\in{\bf K}. That is, q=Ξ»βˆ’1​uβˆ’Ξ»βˆ’1​λ0π‘žsuperscriptπœ†1𝑒superscriptπœ†1subscriptπœ†0q=\lambda^{-1}u-\lambda^{-1}\lambda_{0}. Thus, hβ„Žh is a polynomial of u𝑒u. Then the conclusion follows from Theorem 2.1 in [4].

(2) If degy⁑q=degy⁑hsubscriptdegreeπ‘¦π‘žsubscriptdegreeπ‘¦β„Ž\deg_{y}q=\deg_{y}h, then it follows from the arguments of (1) that u𝑒u is a polynomial of hβ„Žh. It follows from Corollary 2.3 in [13] that u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent. This is a contradiction!

Case III If degy⁑q=9subscriptdegreeπ‘¦π‘ž9\deg_{y}q=9, then degy⁑u=9subscriptdegree𝑦𝑒9\deg_{y}u=9 or degy⁑h=9subscriptdegreeπ‘¦β„Ž9\deg_{y}h=9. That is, degy⁑q=degy⁑usubscriptdegreeπ‘¦π‘žsubscriptdegree𝑦𝑒\deg_{y}q=\deg_{y}u or degy⁑q=degy⁑hsubscriptdegreeπ‘¦π‘žsubscriptdegreeπ‘¦β„Ž\deg_{y}q=\deg_{y}h. Then the conclusion follows from the arguments of Case II. ∎

Theorem 2.5.

Let H=(u​(x,y),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y),v(x,y,z),h(x,y)) be a polynomial map over πŠβ€‹[x,y,z]𝐊π‘₯𝑦𝑧{\bf K}[x,y,z]. Assume that H​(0)=0𝐻00H(0)=0 and the components of H𝐻H are linearly independent over 𝐊𝐊{\bf K}. If J​H𝐽𝐻JH is nilpotent and the variety V​(Qy)𝑉subscript𝑄𝑦V(Q_{y}) is irreducible for any Qy|(uy,hy)conditionalsubscript𝑄𝑦subscript𝑒𝑦subscriptβ„Žπ‘¦Q_{y}|(u_{y},h_{y}) and degy⁑Q|(degy⁑u,degy⁑h)conditionalsubscriptdegree𝑦𝑄subscriptdegree𝑦𝑒subscriptdegreeπ‘¦β„Ž\deg_{y}Q|(\deg_{y}u,\deg_{y}h), then H𝐻H has the form of Theorem 2.2.

Proof.

Let v=vd​zd+β‹―+v1​z+v0𝑣subscript𝑣𝑑superscript𝑧𝑑⋯subscript𝑣1𝑧subscript𝑣0v=v_{d}z^{d}+\cdots+v_{1}z+v_{0}. Then it follows from Lemma 3.2 in [13] that d=1𝑑1d=1 and v1βˆˆπŠβˆ—subscript𝑣1superscript𝐊v_{1}\in{\bf K}^{*}. Since J​H𝐽𝐻JH is nilpotent, so we have the following equations:

{ux+v0​y=0(2.4)ux​v0​yβˆ’v0​x​uyβˆ’v1​hy=0(2.5)v1​(ux​hyβˆ’uy​hx)=0(2.6)\left\{\begin{aligned} u_{x}+v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.4)\\ u_{x}v_{0y}-v_{0x}u_{y}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}(2.5)\\ v_{1}(u_{x}h_{y}-u_{y}h_{x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.6)\end{aligned}\right.

It follows from equation (2.4)2.4(2.4) that ux=βˆ’v0​ysubscript𝑒π‘₯subscript𝑣0𝑦u_{x}=-v_{0y}. Thus, there exists PβˆˆπŠβ€‹[x,y]π‘ƒπŠπ‘₯𝑦P\in{\bf K}[x,y] such that

u=βˆ’Py,v=Px(2.7)formulae-sequence𝑒subscript𝑃𝑦𝑣subscript𝑃π‘₯2.7u=-P_{y},~{}~{}v=P_{x}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.7)

It follows from equation (2.6)2.6(2.6) and Lemma 3.1 in [13] that there exists qβˆˆπŠβ€‹[x,y]π‘žπŠπ‘₯𝑦q\in{\bf K}[x,y] such that

u,hβˆˆπŠβ€‹[q](2.8)formulae-sequenceπ‘’β„ŽπŠdelimited-[]π‘ž2.8u,~{}h\in{\bf K}[q]~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.8)

If degy⁑q≀1subscriptdegreeπ‘¦π‘ž1\deg_{y}q\leq 1, then the conclusion follows from the proof of Theorem 2.8 in [12].

Suppose degy⁑qβ‰₯2subscriptdegreeπ‘¦π‘ž2\deg_{y}q\geq 2. Let q​(x,y)=qt​(x)​yt+qtβˆ’1​(x)​ytβˆ’1+β‹―+q1​(x)​y+q0​(x)π‘žπ‘₯𝑦subscriptπ‘žπ‘‘π‘₯superscript𝑦𝑑subscriptπ‘žπ‘‘1π‘₯superscript𝑦𝑑1β‹―subscriptπ‘ž1π‘₯𝑦subscriptπ‘ž0π‘₯q(x,y)=q_{t}(x)y^{t}+q_{t-1}(x)y^{t-1}+\cdots+q_{1}(x)y+q_{0}(x) with tβ‰₯2𝑑2t\geq 2. It follows from equation (2.8)2.8(2.8) and Lemma 3.4 in [13] that

qt​(x)βˆˆπŠβˆ—(2.30)subscriptπ‘žπ‘‘π‘₯superscript𝐊2.30q_{t}(x)\in{\bf K}^{*}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.30)

It follows from equations (2.5)2.5(2.5), (2.7)2.7(2.7) and (2.8)2.8(2.8) that

qy​[v1​h′​(q)+v0​x​u′​(q)]=βˆ’(u′​(q)​qx)2(2.31)subscriptπ‘žπ‘¦delimited-[]subscript𝑣1superscriptβ„Žβ€²π‘žsubscript𝑣0π‘₯superscriptπ‘’β€²π‘žsuperscriptsuperscriptπ‘’β€²π‘žsubscriptπ‘žπ‘₯22.31q_{y}[v_{1}h^{\prime}(q)+v_{0x}u^{\prime}(q)]=-(u^{\prime}(q)q_{x})^{2}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.31)

Since uy=u′​(q)​qysubscript𝑒𝑦superscriptπ‘’β€²π‘žsubscriptπ‘žπ‘¦u_{y}=u^{\prime}(q)q_{y}, hy=h′​(q)​qysubscriptβ„Žπ‘¦superscriptβ„Žβ€²π‘žsubscriptπ‘žπ‘¦h_{y}=h^{\prime}(q)q_{y}, so we have qy|(uy,hy)conditionalsubscriptπ‘žπ‘¦subscript𝑒𝑦subscriptβ„Žπ‘¦q_{y}|(u_{y},h_{y}) and degy⁑q|(degy⁑u,degy⁑h)conditionalsubscriptdegreeπ‘¦π‘žsubscriptdegree𝑦𝑒subscriptdegreeπ‘¦β„Ž\deg_{y}q|(\deg_{y}u,\deg_{y}h). Thus, the variety V​(qy)𝑉subscriptπ‘žπ‘¦V(q_{y}) is irreducible. That is, there exists an irreducible polynomial f​(x,y)βˆˆπŠβ€‹[x,y]𝑓π‘₯π‘¦πŠπ‘₯𝑦f(x,y)\in{\bf K}[x,y] such that

qy=fs​(x,y)(2.32)subscriptπ‘žπ‘¦superscript𝑓𝑠π‘₯𝑦2.32q_{y}=f^{s}(x,y)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.32)

for some sβˆˆπβˆ—π‘ superscript𝐍s\in{\bf N}^{*}. It follows from equation (2.31)2.31(2.31) that qy|(u′​(q))2​qx2conditionalsubscriptπ‘žπ‘¦superscriptsuperscriptπ‘’β€²π‘ž2subscriptsuperscriptπ‘ž2π‘₯q_{y}|(u^{\prime}(q))^{2}q^{2}_{x}. That is, fs​(x,y)|(u′​(q))2​qx2conditionalsuperscript𝑓𝑠π‘₯𝑦superscriptsuperscriptπ‘’β€²π‘ž2subscriptsuperscriptπ‘ž2π‘₯f^{s}(x,y)|(u^{\prime}(q))^{2}q^{2}_{x}. Since f​(x,y)𝑓π‘₯𝑦f(x,y) is irreducible, so we have f​(x,y)|u′​(q)conditional𝑓π‘₯𝑦superscriptπ‘’β€²π‘žf(x,y)|u^{\prime}(q) or f​(x,y)|qxconditional𝑓π‘₯𝑦subscriptπ‘žπ‘₯f(x,y)|q_{x}.

Case I Suppose f​(x,y)|u′​(q)conditional𝑓π‘₯𝑦superscriptπ‘’β€²π‘žf(x,y)|u^{\prime}(q). Since u′​(q)superscriptπ‘’β€²π‘žu^{\prime}(q) is a polynomial of qπ‘žq, so it follows from the Fundamental Theorem of Algebra that u′​(q)=c0​(q+c1)​(q+c2)​⋯​(q+ck)superscriptπ‘’β€²π‘žsubscript𝑐0π‘žsubscript𝑐1π‘žsubscript𝑐2β‹―π‘žsubscriptπ‘π‘˜u^{\prime}(q)=c_{0}(q+c_{1})(q+c_{2})\cdots(q+c_{k}) for c0βˆˆπŠβˆ—subscript𝑐0superscript𝐊c_{0}\in{\bf K}^{*} and ci∈𝐊subscriptπ‘π‘–πŠc_{i}\in{\bf K}, 1≀i≀k1π‘–π‘˜1\leq i\leq k. If f​(x,y)|u′​(q)conditional𝑓π‘₯𝑦superscriptπ‘’β€²π‘žf(x,y)|u^{\prime}(q), then there exists i0∈{1,2,…,k}subscript𝑖012β€¦π‘˜i_{0}\in\{1,2,\ldots,k\} such that f​(x,y)|(q+ci0)conditional𝑓π‘₯π‘¦π‘žsubscript𝑐subscript𝑖0f(x,y)|(q+c_{i_{0}}). That is,

q+ci0=f​(x,y)​M(1)​(x,y)(2.33)π‘žsubscript𝑐subscript𝑖0𝑓π‘₯𝑦superscript𝑀1π‘₯𝑦2.33q+c_{i_{0}}=f(x,y)M^{(1)}(x,y)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.33)

for some M(1)​(x,y)βˆˆπŠβ€‹[x,y]superscript𝑀1π‘₯π‘¦πŠπ‘₯𝑦M^{(1)}(x,y)\in{\bf K}[x,y]. Thus, it follows from equations (2.33)2.33(2.33) and (2.32)2.32(2.32) that qy=fy​M(1)+f​My(1)=fssubscriptπ‘žπ‘¦subscript𝑓𝑦superscript𝑀1𝑓subscriptsuperscript𝑀1𝑦superscript𝑓𝑠q_{y}=f_{y}M^{(1)}+fM^{(1)}_{y}=f^{s}. That is,

fy​M(1)=f​(fsβˆ’1βˆ’My(1)).subscript𝑓𝑦superscript𝑀1𝑓superscript𝑓𝑠1subscriptsuperscript𝑀1𝑦f_{y}M^{(1)}=f(f^{s-1}-M^{(1)}_{y}).

Since f𝑓f is irreducible and fyβ‰ 0subscript𝑓𝑦0f_{y}\neq 0, so it follows from the above equation that f|M(1)conditional𝑓superscript𝑀1f|M^{(1)}. That is,

M(1)​(x,y)=f​(x,y)​M(2)​(x,y)(2.34)superscript𝑀1π‘₯𝑦𝑓π‘₯𝑦superscript𝑀2π‘₯𝑦2.34M^{(1)}(x,y)=f(x,y)M^{(2)}(x,y)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.34)

for some M(2)​(x,y)βˆˆπŠβ€‹[x,y]superscript𝑀2π‘₯π‘¦πŠπ‘₯𝑦M^{(2)}(x,y)\in{\bf K}[x,y]. Thus, it follows from equations (2.33)2.33(2.33) and (2.34)2.34(2.34) that

q+ci0=f2​(x,y)​M(2)​(x,y)(2.35)π‘žsubscript𝑐subscript𝑖0superscript𝑓2π‘₯𝑦superscript𝑀2π‘₯𝑦2.35q+c_{i_{0}}=f^{2}(x,y)M^{(2)}(x,y)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.35)

Then it follows from equation (2.35)2.35(2.35) that qy=2​f​fy​M(2)+f2​My(2)=fssubscriptπ‘žπ‘¦2𝑓subscript𝑓𝑦superscript𝑀2superscript𝑓2subscriptsuperscript𝑀2𝑦superscript𝑓𝑠q_{y}=2ff_{y}M^{(2)}+f^{2}M^{(2)}_{y}=f^{s}. That is,

2​fy​M(2)=f​(fsβˆ’2βˆ’My(2)).2subscript𝑓𝑦superscript𝑀2𝑓superscript𝑓𝑠2subscriptsuperscript𝑀2𝑦2f_{y}M^{(2)}=f(f^{s-2}-M^{(2)}_{y}).

Since f𝑓f is irreducible and fyβ‰ 0subscript𝑓𝑦0f_{y}\neq 0, so we have f|M(2)conditional𝑓superscript𝑀2f|M^{(2)}. We can do step by step until

q+ci0=fs+1​M(s+1)π‘žsubscript𝑐subscript𝑖0superscript𝑓𝑠1superscript𝑀𝑠1q+c_{i_{0}}=f^{s+1}M^{(s+1)}

for some M(s+1)βˆˆπŠβ€‹[x,y]superscript𝑀𝑠1𝐊π‘₯𝑦M^{(s+1)}\in{\bf K}[x,y]. Then it follows from the above equation and equation (2.32)2.32(2.32) that qy=(s+1)​fs​fy​M(s+1)+fs+1​My(s+1)=fssubscriptπ‘žπ‘¦π‘ 1superscript𝑓𝑠subscript𝑓𝑦superscript𝑀𝑠1superscript𝑓𝑠1subscriptsuperscript𝑀𝑠1𝑦superscript𝑓𝑠q_{y}=(s+1)f^{s}f_{y}M^{(s+1)}+f^{s+1}M^{(s+1)}_{y}=f^{s}. That is,

(s+1)​fy​M(s+1)+f​My(s+1)=1(2.36)𝑠1subscript𝑓𝑦superscript𝑀𝑠1𝑓subscriptsuperscript𝑀𝑠1𝑦12.36(s+1)f_{y}M^{(s+1)}+fM^{(s+1)}_{y}=1~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.36)

Let f​(x,y)=fl​(x)​yl+flβˆ’1​(x)​ylβˆ’1+β‹―+f1​(x)​y+f0​(x)𝑓π‘₯𝑦subscript𝑓𝑙π‘₯superscript𝑦𝑙subscript𝑓𝑙1π‘₯superscript𝑦𝑙1β‹―subscript𝑓1π‘₯𝑦subscript𝑓0π‘₯f(x,y)=f_{l}(x)y^{l}+f_{l-1}(x)y^{l-1}+\cdots+f_{1}(x)y+f_{0}(x), M(s+1)=Mm(s+1)​(x)​ym+Mmβˆ’1(s+1)​(x)​ymβˆ’1+β‹―+M1(s+1)​(x)​y+M0(s+1)​(x)superscript𝑀𝑠1subscriptsuperscript𝑀𝑠1π‘šπ‘₯superscriptπ‘¦π‘šsubscriptsuperscript𝑀𝑠1π‘š1π‘₯superscriptπ‘¦π‘š1β‹―subscriptsuperscript𝑀𝑠11π‘₯𝑦subscriptsuperscript𝑀𝑠10π‘₯M^{(s+1)}=M^{(s+1)}_{m}(x)y^{m}+M^{(s+1)}_{m-1}(x)y^{m-1}+\cdots+M^{(s+1)}_{1}(x)y+M^{(s+1)}_{0}(x) with fl​(x)​Mm(s+1)​(x)β‰ 0subscript𝑓𝑙π‘₯subscriptsuperscript𝑀𝑠1π‘šπ‘₯0f_{l}(x)M^{(s+1)}_{m}(x)\neq 0 and lβ‰₯1𝑙1l\geq 1. It follows from equations (2.32)2.32(2.32) and (2.30)2.30(2.30) that fls​(x)=t​qt​(x)βˆˆπŠβˆ—subscriptsuperscript𝑓𝑠𝑙π‘₯𝑑subscriptπ‘žπ‘‘π‘₯superscript𝐊f^{s}_{l}(x)=tq_{t}(x)\in{\bf K}^{*}. That is, fl​(x)βˆˆπŠβˆ—subscript𝑓𝑙π‘₯superscript𝐊f_{l}(x)\in{\bf K}^{*}. It follows from equation (2.36)2.36(2.36) that (s+1)​(l​fl​ylβˆ’1+(lβˆ’1)​flβˆ’1​(x)​ylβˆ’2+β‹―+f1​(x))​(Mm(s+1)​(x)​ym+Mmβˆ’1(s+1)​(x)​ymβˆ’1+β‹―+M1(s+1)​(x)​y+M0(s+1)​(x))+(fl​(x)​yl+flβˆ’1​(x)​ylβˆ’1+β‹―+f1​(x)​y+f0​(x))​(m​Mm(s+1)​(x)​ymβˆ’1+(mβˆ’1)​Mmβˆ’1(s+1)​(x)​ymβˆ’2+β‹―+M1(s+1)​(x))=1𝑠1𝑙subscript𝑓𝑙superscript𝑦𝑙1𝑙1subscript𝑓𝑙1π‘₯superscript𝑦𝑙2β‹―subscript𝑓1π‘₯subscriptsuperscript𝑀𝑠1π‘šπ‘₯superscriptπ‘¦π‘šsubscriptsuperscript𝑀𝑠1π‘š1π‘₯superscriptπ‘¦π‘š1β‹―subscriptsuperscript𝑀𝑠11π‘₯𝑦subscriptsuperscript𝑀𝑠10π‘₯subscript𝑓𝑙π‘₯superscript𝑦𝑙subscript𝑓𝑙1π‘₯superscript𝑦𝑙1β‹―subscript𝑓1π‘₯𝑦subscript𝑓0π‘₯π‘šsubscriptsuperscript𝑀𝑠1π‘šπ‘₯superscriptπ‘¦π‘š1π‘š1subscriptsuperscript𝑀𝑠1π‘š1π‘₯superscriptπ‘¦π‘š2β‹―subscriptsuperscript𝑀𝑠11π‘₯1(s+1)(lf_{l}y^{l-1}+(l-1)f_{l-1}(x)y^{l-2}+\cdots+f_{1}(x))(M^{(s+1)}_{m}(x)y^{m}+M^{(s+1)}_{m-1}(x)y^{m-1}+\cdots+M^{(s+1)}_{1}(x)y+M^{(s+1)}_{0}(x))+(f_{l}(x)y^{l}+f_{l-1}(x)y^{l-1}+\cdots+f_{1}(x)y+f_{0}(x))(mM^{(s+1)}_{m}(x)y^{m-1}+(m-1)M^{(s+1)}_{m-1}(x)y^{m-2}+\cdots+M^{(s+1)}_{1}(x))=1Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (2.37)
We always view that the polynomials are in πŠβ€‹[x]​[y]𝐊delimited-[]π‘₯delimited-[]𝑦{\bf K}[x][y] with coefficients in πŠβ€‹[x]𝐊delimited-[]π‘₯{\bf K}[x] when comparing the coefficients of yjsuperscript𝑦𝑗y^{j}.

Suppose mβ‰₯1π‘š1m\geq 1. Comparing the coefficients of yl+mβˆ’1superscriptπ‘¦π‘™π‘š1y^{l+m-1} of equation (2.37)2.37(2.37), we have the following equation:

(s+1)​l​fl​Mm(s+1)​(x)+m​fl​Mm(s+1)​(x)=0𝑠1𝑙subscript𝑓𝑙subscriptsuperscript𝑀𝑠1π‘šπ‘₯π‘šsubscript𝑓𝑙subscriptsuperscript𝑀𝑠1π‘šπ‘₯0(s+1)lf_{l}M^{(s+1)}_{m}(x)+mf_{l}M^{(s+1)}_{m}(x)=0

Since fl​(x)​Mm(s+1)​(x)β‰ 0subscript𝑓𝑙π‘₯subscriptsuperscript𝑀𝑠1π‘šπ‘₯0f_{l}(x)M^{(s+1)}_{m}(x)\neq 0, so we have (s+1)​l+m=0𝑠1π‘™π‘š0(s+1)l+m=0. This is a contradiction because s​lβ‰₯1𝑠𝑙1sl\geq 1 and mβ‰₯1π‘š1m\geq 1. Therefore, we have m=0π‘š0m=0. That is, My(s+1)=0subscriptsuperscript𝑀𝑠1𝑦0M^{(s+1)}_{y}=0. Then equation (2.36)2.36(2.36) has the following form:

(s+1)​fy​M(s+1)=1𝑠1subscript𝑓𝑦superscript𝑀𝑠11(s+1)f_{y}M^{(s+1)}=1

Thus, we have fyβˆˆπŠβˆ—subscript𝑓𝑦superscript𝐊f_{y}\in{\bf K}^{*}. That is, f=f1​y+f0​(x)𝑓subscript𝑓1𝑦subscript𝑓0π‘₯f=f_{1}y+f_{0}(x) with f1βˆˆπŠβˆ—subscript𝑓1superscript𝐊f_{1}\in{\bf K}^{*}. So f𝑓f is a polynomial of y+a​(x)π‘¦π‘Žπ‘₯y+a(x) with a​(x)=f1βˆ’1​f0​(x)π‘Žπ‘₯superscriptsubscript𝑓11subscript𝑓0π‘₯a(x)=f_{1}^{-1}f_{0}(x). That is, q,u,hβˆˆπŠβ€‹[y+a​(x)]π‘žπ‘’β„ŽπŠdelimited-[]π‘¦π‘Žπ‘₯q,~{}u,~{}h\in{\bf K}[y+a(x)]. Then the conclusion follows from the proof of Theorem 2.8 in [12].

Case II If f​(x,y)|qxconditional𝑓π‘₯𝑦subscriptπ‘žπ‘₯f(x,y)|q_{x}, then

qx=f​(x,y)​N(1)​(x,y)(2.38)subscriptπ‘žπ‘₯𝑓π‘₯𝑦superscript𝑁1π‘₯𝑦2.38q_{x}=f(x,y)N^{(1)}(x,y)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.38)

for some N(1)​(x,y)βˆˆπŠβ€‹[x,y]superscript𝑁1π‘₯π‘¦πŠπ‘₯𝑦N^{(1)}(x,y)\in{\bf K}[x,y]. Thus, it follows from equations (2.38)2.38(2.38) and (2.32)2.32(2.32) that qx​y=fy​N(1)+f​Ny(1)subscriptπ‘žπ‘₯𝑦subscript𝑓𝑦superscript𝑁1𝑓subscriptsuperscript𝑁1𝑦q_{xy}=f_{y}N^{(1)}+fN^{(1)}_{y}, qy​x=s​fsβˆ’1​fxsubscriptπ‘žπ‘¦π‘₯𝑠superscript𝑓𝑠1subscript𝑓π‘₯q_{yx}=sf^{s-1}f_{x}. Thus, we have the following equation:

fy​N(1)=f​(s​fsβˆ’2​fxβˆ’Ny(1)).subscript𝑓𝑦superscript𝑁1𝑓𝑠superscript𝑓𝑠2subscript𝑓π‘₯subscriptsuperscript𝑁1𝑦f_{y}N^{(1)}=f(sf^{s-2}f_{x}-N^{(1)}_{y}).

Since f𝑓f is irreducible and fyβ‰ 0subscript𝑓𝑦0f_{y}\neq 0, so we have f|N(1)conditional𝑓superscript𝑁1f|N^{(1)}. That is, N(1)=f​N(2)superscript𝑁1𝑓superscript𝑁2N^{(1)}=fN^{(2)} for some N(2)βˆˆπŠβ€‹[x,y]superscript𝑁2𝐊π‘₯𝑦N^{(2)}\in{\bf K}[x,y]. It follows from equation (2.38)2.38(2.38) that qx=f2​N(2)​(x,y)subscriptπ‘žπ‘₯superscript𝑓2superscript𝑁2π‘₯𝑦q_{x}=f^{2}N^{(2)}(x,y). We can do step by step until qx=fs​N(s)​(x,y)subscriptπ‘žπ‘₯superscript𝑓𝑠superscript𝑁𝑠π‘₯𝑦q_{x}=f^{s}N^{(s)}(x,y). Then qx​y=s​fsβˆ’1​fy​N(s)+fs​Ny(s)=qy​x=s​fsβˆ’1​fxsubscriptπ‘žπ‘₯𝑦𝑠superscript𝑓𝑠1subscript𝑓𝑦superscript𝑁𝑠superscript𝑓𝑠subscriptsuperscript𝑁𝑠𝑦subscriptπ‘žπ‘¦π‘₯𝑠superscript𝑓𝑠1subscript𝑓π‘₯q_{xy}=sf^{s-1}f_{y}N^{(s)}+f^{s}N^{(s)}_{y}=q_{yx}=sf^{s-1}f_{x}. That is,

s​(fxβˆ’fy​N(s))=f​Ny(s)(2.39)𝑠subscript𝑓π‘₯subscript𝑓𝑦superscript𝑁𝑠𝑓subscriptsuperscript𝑁𝑠𝑦2.39s(f_{x}-f_{y}N^{(s)})=fN^{(s)}_{y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.39)

Since qy=fssubscriptπ‘žπ‘¦superscript𝑓𝑠q_{y}=f^{s}, qx=fs​N(s)​(x,y)subscriptπ‘žπ‘₯superscript𝑓𝑠superscript𝑁𝑠π‘₯𝑦q_{x}=f^{s}N^{(s)}(x,y) and degy⁑qx≀tβˆ’1=degy⁑qysubscriptdegree𝑦subscriptπ‘žπ‘₯𝑑1subscriptdegree𝑦subscriptπ‘žπ‘¦\deg_{y}q_{x}\leq t-1=\deg_{y}q_{y}, so we have degy⁑(fs​N(s)​(x,y))≀degy⁑(fs)subscriptdegree𝑦superscript𝑓𝑠superscript𝑁𝑠π‘₯𝑦subscriptdegree𝑦superscript𝑓𝑠\deg_{y}(f^{s}N^{(s)}(x,y))\leq\deg_{y}(f^{s}). Thus, we have degy⁑N(s)​(x,y)=0subscriptdegree𝑦superscript𝑁𝑠π‘₯𝑦0\deg_{y}N^{(s)}(x,y)=0. That is, Ny(s)=0subscriptsuperscript𝑁𝑠𝑦0N^{(s)}_{y}=0. Then equation (2.39)2.39(2.39) has the following form:

fx=fy​N(s)​(x)(2.40)subscript𝑓π‘₯subscript𝑓𝑦superscript𝑁𝑠π‘₯2.40f_{x}=f_{y}N^{(s)}(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.40)

where N(s)​(x):=N(s)​(x,0)=N(s)​(x,y)assignsuperscript𝑁𝑠π‘₯superscript𝑁𝑠π‘₯0superscript𝑁𝑠π‘₯𝑦N^{(s)}(x):=N^{(s)}(x,0)=N^{(s)}(x,y). Let xΒ―=xΒ―π‘₯π‘₯\bar{x}=x, yΒ―=y+∫N(s)​(x)​𝑑x¯𝑦𝑦superscript𝑁𝑠π‘₯differential-dπ‘₯\bar{y}=y+\int N^{(s)}(x)dx. Then it follows from equation (2.40)2.40(2.40) that fxΒ―=0subscript𝑓¯π‘₯0f_{\bar{x}}=0. That is, fβˆˆπŠβ€‹[y+a​(x)]π‘“πŠdelimited-[]π‘¦π‘Žπ‘₯f\in{\bf K}[y+a(x)], where a​(x)=∫N(s)​(x)​𝑑xβˆˆπŠβ€‹[x]π‘Žπ‘₯superscript𝑁𝑠π‘₯differential-dπ‘₯𝐊delimited-[]π‘₯a(x)=\int N^{(s)}(x)dx\in{\bf K}[x]. Thus, q,u,hπ‘žπ‘’β„Žq,~{}u,~{}h are polynomials of y+a​(x)π‘¦π‘Žπ‘₯y+a(x). Then the conclusion follows from the proof of Theorem 2.8 in [12]. ∎

Corollary 2.6.

Let H=(u​(x,y),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y),v(x,y,z),h(x,y)) be a polynomial map over πŠβ€‹[x,y,z]𝐊π‘₯𝑦𝑧{\bf K}[x,y,z]. Assume that H​(0)=0𝐻00H(0)=0 and the components of H𝐻H are linearly independent over 𝐊𝐊{\bf K}. If J​H𝐽𝐻JH is nilpotent and the variety V​(uy)𝑉subscript𝑒𝑦V(u_{y}) or the variety V​(hy)𝑉subscriptβ„Žπ‘¦V(h_{y}) is irreducible, then H𝐻H has the form of Theorem 2.2.

Proof.

Let v=vd​zd+β‹―+v1​z+v0𝑣subscript𝑣𝑑superscript𝑧𝑑⋯subscript𝑣1𝑧subscript𝑣0v=v_{d}z^{d}+\cdots+v_{1}z+v_{0}. Then it follows from Lemma 3.2 in [13] that d=1𝑑1d=1 and v1βˆˆπŠβˆ—subscript𝑣1superscript𝐊v_{1}\in{\bf K}^{*}. Since J​H𝐽𝐻JH is nilpotent, so we have the following equations:

{ux+v0​y=0(2.4)ux​v0​yβˆ’v0​x​uyβˆ’v1​hy=0(2.5)v1​(ux​hyβˆ’uy​hx)=0(2.6)\left\{\begin{aligned} u_{x}+v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.4)\\ u_{x}v_{0y}-v_{0x}u_{y}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}(2.5)\\ v_{1}(u_{x}h_{y}-u_{y}h_{x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.6)\end{aligned}\right.

It follows from equation (2.6)2.6(2.6) and Lemma 3.1 in [13] that there exists qβˆˆπŠβ€‹[x,y]π‘žπŠπ‘₯𝑦q\in{\bf K}[x,y] such that

u,hβˆˆπŠβ€‹[q](2.8)formulae-sequenceπ‘’β„ŽπŠdelimited-[]π‘ž2.8u,~{}h\in{\bf K}[q]~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2.8)

Since V​(uy)𝑉subscript𝑒𝑦V(u_{y}) or V​(hy)𝑉subscriptβ„Žπ‘¦V(h_{y}) is irreducible and V​(qy)βŠ‚V​(uy)∩V​(hy)𝑉subscriptπ‘žπ‘¦π‘‰subscript𝑒𝑦𝑉subscriptβ„Žπ‘¦V(q_{y})\subset V(u_{y})\cap V(h_{y}) is a subvariety, so V​(qy)𝑉subscriptπ‘žπ‘¦V(q_{y}) is irreducible. Then the conclusion follows from the proof of Theorem 2.5. ∎

3 Polynomial maps of the form H=(u​(x,y,z),v​(x,y,u),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦π‘’β„Žπ‘₯𝑦H=(u(x,y,z),v(x,\allowbreak y,u),h(x,y))

In this section, we classify polynomial maps of the form H=(u​(x,y,z),v​(x,y,u),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦π‘’β„Žπ‘₯𝑦H=(u(x,y,z),v(x,y,u),\allowbreak h(x,y)) in the case that J​H𝐽𝐻JH is nilpotent and (deg⁑v​(x,y,0),deg⁑h)≀3degree𝑣π‘₯𝑦0degreeβ„Ž3(\deg v(x,y,0),\deg h)\leq 3. Combining Theorem 2.2 with Corollary 3.7 in [13], we have the following Proposition.

Proposition 3.1.

Let H=(u​(x,y,z),v​(x,y),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦β„Žπ‘₯𝑦H=(u(x,y,z),v(x,y),h(x,y)) be a polynomial map over πŠβ€‹[x,y,z]𝐊π‘₯𝑦𝑧{\bf K}[x,y,z]. Assume that H​(0)=0𝐻00H(0)=0 and the components of H𝐻H are linearly independent over 𝐊𝐊{\bf K}. If J​H𝐽𝐻JH is nilpotent and (degx⁑v,degx⁑h)≀3subscriptdegreeπ‘₯𝑣subscriptdegreeπ‘₯β„Ž3(\deg_{x}v,\deg_{x}h)\leq 3 or at least one of degx⁑vsubscriptdegreeπ‘₯𝑣\deg_{x}v, degx⁑hsubscriptdegreeπ‘₯β„Ž\deg_{x}h is a prime, then u=u1​zβˆ’aβˆ’1​b′​(y)​g​(a​x+b​(y))βˆ’u1​l2​y𝑒subscript𝑒1𝑧superscriptπ‘Ž1superscriptπ‘β€²π‘¦π‘”π‘Žπ‘₯𝑏𝑦subscript𝑒1subscript𝑙2𝑦u=u_{1}z-a^{-1}b^{\prime}(y)g(ax+b(y))-u_{1}l_{2}y, v=g​(a​x+b​(y))π‘£π‘”π‘Žπ‘₯𝑏𝑦v=g(ax+b(y)), h=c0​v2+l2​vβ„Žsubscript𝑐0superscript𝑣2subscript𝑙2𝑣h=c_{0}v^{2}+l_{2}v, where b​(y)=u1​c0​a​y2+l1​y+l~2𝑏𝑦subscript𝑒1subscript𝑐0π‘Žsuperscript𝑦2subscript𝑙1𝑦subscript~𝑙2b(y)=u_{1}c_{0}ay^{2}+l_{1}y+\tilde{l}_{2}, u1,c0,aβˆˆπŠβˆ—subscript𝑒1subscript𝑐0π‘Žsuperscript𝐊u_{1},c_{0},a\in{\bf K}^{*}, l1,l2,l~2∈𝐊subscript𝑙1subscript𝑙2subscript~𝑙2𝐊l_{1},l_{2},\tilde{l}_{2}\in{\bf K}, g​(t)βˆˆπŠβ€‹[t]π‘”π‘‘πŠdelimited-[]𝑑g(t)\in{\bf K}[t] and g​(0)=0𝑔00g(0)=0, degt⁑g​(t)β‰₯1subscriptdegree𝑑𝑔𝑑1\deg_{t}g(t)\geq 1.

In the following theorem, we denote vx=βˆ‚vβˆ‚x​(x,y,u)subscript𝑣π‘₯𝑣π‘₯π‘₯𝑦𝑒v_{x}=\frac{\partial v}{\partial x}(x,y,u), vy=βˆ‚vβˆ‚y​(x,y,u)subscript𝑣𝑦𝑣𝑦π‘₯𝑦𝑒v_{y}=\frac{\partial v}{\partial y}(x,y,u) and vu=βˆ‚vβˆ‚u​(x,y,u)subscript𝑣𝑒𝑣𝑒π‘₯𝑦𝑒v_{u}=\frac{\partial v}{\partial u}(x,y,u).

Theorem 3.2.

Let H=(u​(x,y,z),v​(x,y,u),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦π‘’β„Žπ‘₯𝑦H=(u(x,y,z),v(x,y,u),h(x,y)) be a polynomial map over πŠβ€‹[x,y,z]𝐊π‘₯𝑦𝑧{\bf K}[x,y,z]. Assume that H​(0)=0𝐻00H(0)=0 and degu⁑vβ‰₯2subscriptdegree𝑒𝑣2\deg_{u}v\geq 2. If J​H𝐽𝐻JH is nilpotent, then u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.

Proof.

If uz=0subscript𝑒𝑧0u_{z}=0, then it follows from Proposition 2.1 in [12] that u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent. Thus, we can assume that uzβ‰ 0subscript𝑒𝑧0u_{z}\neq 0. Since J​H𝐽𝐻JH is nilpotent, we have the following equations:

{ux+vy=βˆ’vu​uy(3.1)ux​vyβˆ’vx​uyβˆ’hx​uz=hy​vu​uz(3.2)uz​(vx​hyβˆ’vy​hx)=0(3.3)\left\{\begin{aligned} u_{x}+v_{y}=-v_{u}u_{y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.1)\\ u_{x}v_{y}-v_{x}u_{y}-h_{x}u_{z}=h_{y}v_{u}u_{z}~{}~{}~{}~{}~{}~{}~{}~{}(3.2)\\ u_{z}(v_{x}h_{y}-v_{y}h_{x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.3)\end{aligned}\right.

Let u=ud​zd+udβˆ’1​zdβˆ’1+β‹―+u1​z+u0𝑒subscript𝑒𝑑superscript𝑧𝑑subscript𝑒𝑑1superscript𝑧𝑑1β‹―subscript𝑒1𝑧subscript𝑒0u=u_{d}z^{d}+u_{d-1}z^{d-1}+\cdots+u_{1}z+u_{0}, v​(x,y,u)=vm​(x,y)​um+vmβˆ’1​(x,y)​umβˆ’1+β‹―+v1​(x,y)​u+v0​(x,y)𝑣π‘₯𝑦𝑒subscriptπ‘£π‘šπ‘₯𝑦superscriptπ‘’π‘šsubscriptπ‘£π‘š1π‘₯𝑦superscriptπ‘’π‘š1β‹―subscript𝑣1π‘₯𝑦𝑒subscript𝑣0π‘₯𝑦v(x,y,u)=v_{m}(x,y)u^{m}+v_{m-1}(x,y)u^{m-1}+\cdots+v_{1}(x,y)u+v_{0}(x,y) with ud​vmβ‰ 0subscript𝑒𝑑subscriptπ‘£π‘š0u_{d}v_{m}\neq 0 and dβ‰₯1𝑑1d\geq 1, mβ‰₯2π‘š2m\geq 2, ui,vjβˆˆπŠβ€‹[x,y]subscript𝑒𝑖subscriptπ‘£π‘—πŠπ‘₯𝑦u_{i},v_{j}\in{\bf K}[x,y] for 0≀i≀d0𝑖𝑑0\leq i\leq d, 0≀j≀m0π‘—π‘š0\leq j\leq m. It follows from equation (3.1)3.1(3.1) that ud​x​zd+u(dβˆ’1)​x​zdβˆ’1+β‹―+u1​x​z+u0​x+vm​y​um+v(mβˆ’1)​y​umβˆ’1+β‹―+v1​y​u+v0​y=βˆ’(m​vm​umβˆ’1+β‹―+2​v2​u+v1)​(ud​y​zd+u(dβˆ’1)​y​zdβˆ’1+β‹―+u1​y​z+u0​y)subscript𝑒𝑑π‘₯superscript𝑧𝑑subscript𝑒𝑑1π‘₯superscript𝑧𝑑1β‹―subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscriptπ‘£π‘šπ‘¦superscriptπ‘’π‘šsubscriptπ‘£π‘š1𝑦superscriptπ‘’π‘š1β‹―subscript𝑣1𝑦𝑒subscript𝑣0π‘¦π‘šsubscriptπ‘£π‘šsuperscriptπ‘’π‘š1β‹―2subscript𝑣2𝑒subscript𝑣1subscript𝑒𝑑𝑦superscript𝑧𝑑subscript𝑒𝑑1𝑦superscript𝑧𝑑1β‹―subscript𝑒1𝑦𝑧subscript𝑒0𝑦u_{dx}z^{d}+u_{(d-1)x}z^{d-1}+\cdots+u_{1x}z+u_{0x}+v_{my}u^{m}+v_{(m-1)y}u^{m-1}+\cdots+v_{1y}u+v_{0y}=-(mv_{m}u^{m-1}+\cdots+2v_{2}u+v_{1})(u_{dy}z^{d}+u_{(d-1)y}z^{d-1}+\cdots+u_{1y}z+u_{0y})Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.4)
We always view that the polynomials are in πŠβ€‹[x,y]​[z]𝐊π‘₯𝑦delimited-[]𝑧{\bf K}[x,y][z] with coefficients in πŠβ€‹[x,y]𝐊π‘₯𝑦{\bf K}[x,y] in the following arguments.

Since mβ‰₯2π‘š2m\geq 2, comparing the coefficients of zm​dsuperscriptπ‘§π‘šπ‘‘z^{md} of equation (3.4)3.4(3.4), we have vm​y​ud+m​vm​ud​y=0subscriptπ‘£π‘šπ‘¦subscriptπ‘’π‘‘π‘šsubscriptπ‘£π‘šsubscript𝑒𝑑𝑦0v_{my}u_{d}+mv_{m}u_{dy}=0. That is,

vm​yvm=βˆ’m​ud​yudsubscriptπ‘£π‘šπ‘¦subscriptπ‘£π‘šπ‘šsubscript𝑒𝑑𝑦subscript𝑒𝑑\frac{v_{my}}{v_{m}}=-m\frac{u_{dy}}{u_{d}}

Suppose vm​yβ‰ 0subscriptπ‘£π‘šπ‘¦0v_{my}\neq 0. Then ud​yβ‰ 0subscript𝑒𝑑𝑦0u_{dy}\neq 0. Thus, we have vm​udm=ec​(x)subscriptπ‘£π‘šsubscriptsuperscriptπ‘’π‘šπ‘‘superscript𝑒𝑐π‘₯v_{m}u^{m}_{d}=e^{c(x)} by integrating the two sides of the above equation with respect to y𝑦y, where c​(x)𝑐π‘₯c(x) is a function of xπ‘₯x. Since vm,udβˆˆπŠβ€‹[x,y]subscriptπ‘£π‘šsubscriptπ‘’π‘‘πŠπ‘₯𝑦v_{m}~{},u_{d}\in{\bf K}[x,y] and ec​(x)superscript𝑒𝑐π‘₯e^{c(x)} is a function of xπ‘₯x. Thus, we have vm,udβˆˆπŠβ€‹[x]subscriptπ‘£π‘šsubscriptπ‘’π‘‘πŠdelimited-[]π‘₯v_{m}~{},u_{d}\in{\bf K}[x]. This is a contradiction! Therefore, we have vm​y=0=ud​ysubscriptπ‘£π‘šπ‘¦0subscript𝑒𝑑𝑦v_{my}=0=u_{dy}. Then we have ui​y=0subscript𝑒𝑖𝑦0u_{iy}=0 by comparing the coefficients of zd​(mβˆ’1)+isuperscriptπ‘§π‘‘π‘š1𝑖z^{d(m-1)+i} of equation (3.4)3.4(3.4) for i=dβˆ’1,dβˆ’2,…,1𝑖𝑑1𝑑2…1i=d-1,~{}d-2,\ldots,1. Then equation (3.4)3.4(3.4) has the following form:
ud​x​zd+u(dβˆ’1)​x​zdβˆ’1+β‹―+u1​x​z+u0​x+v(mβˆ’1)​y​umβˆ’1+β‹―+v1​y​u+v0​y=βˆ’(m​vm​umβˆ’1+β‹―+2​v2​u+v1)​u0​ysubscript𝑒𝑑π‘₯superscript𝑧𝑑subscript𝑒𝑑1π‘₯superscript𝑧𝑑1β‹―subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscriptπ‘£π‘š1𝑦superscriptπ‘’π‘š1β‹―subscript𝑣1𝑦𝑒subscript𝑣0π‘¦π‘šsubscriptπ‘£π‘šsuperscriptπ‘’π‘š1β‹―2subscript𝑣2𝑒subscript𝑣1subscript𝑒0𝑦u_{dx}z^{d}+u_{(d-1)x}z^{d-1}+\cdots+u_{1x}z+u_{0x}+v_{(m-1)y}u^{m-1}+\cdots+v_{1y}u+v_{0y}=-(mv_{m}u^{m-1}+\cdots+2v_{2}u+v_{1})u_{0y}Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.5)
Comparing the coefficients of zj​dsuperscript𝑧𝑗𝑑z^{jd} for j=mβˆ’1,mβˆ’2,…,2π‘—π‘š1π‘š2…2j=m-1,~{}m-2,\ldots,2 of equation (3.5)3.5(3.5), we have the following equations:

vj​y+(j+1)​vj+1​u0​y=0(3.6)subscript𝑣𝑗𝑦𝑗1subscript𝑣𝑗1subscript𝑒0𝑦03.6v_{jy}+(j+1)v_{j+1}u_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.6)

Then we have

(v1​y+2​v2​u0​y)​uk=βˆ’uk​x(3.7)subscript𝑣1𝑦2subscript𝑣2subscript𝑒0𝑦subscriptπ‘’π‘˜subscriptπ‘’π‘˜π‘₯3.7(v_{1y}+2v_{2}u_{0y})u_{k}=-u_{kx}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.7)

by comparing the coefficients of zksuperscriptπ‘§π‘˜z^{k} of equation (3.5)3.5(3.5) for k=d,dβˆ’1,…,1π‘˜π‘‘π‘‘1…1k=d,~{}d-1,\ldots,1. If v1​y+2​v2​u0​y=0subscript𝑣1𝑦2subscript𝑣2subscript𝑒0𝑦0v_{1y}+2v_{2}u_{0y}=0, then uk​x=0subscriptπ‘’π‘˜π‘₯0u_{kx}=0 for 1≀k≀d1π‘˜π‘‘1\leq k\leq d. If v1​y+2​v2​u0​yβ‰ 0subscript𝑣1𝑦2subscript𝑣2subscript𝑒0𝑦0v_{1y}+2v_{2}u_{0y}\neq 0, then we have uk​x=0subscriptπ‘’π‘˜π‘₯0u_{kx}=0 by comparing the degree of xπ‘₯x of two sides of equation (3.7)3.7(3.7) for 1≀k≀d1π‘˜π‘‘1\leq k\leq d. Since udβ‰ 0subscript𝑒𝑑0u_{d}\neq 0, we have v1​y+2​v2​u0​y=0subscript𝑣1𝑦2subscript𝑣2subscript𝑒0𝑦0v_{1y}+2v_{2}u_{0y}=0. This is a contradiction! Thus, we have v1​y+2​v2​u0​y=0subscript𝑣1𝑦2subscript𝑣2subscript𝑒0𝑦0v_{1y}+2v_{2}u_{0y}=0 and ud​x=β‹―=u1​x=0subscript𝑒𝑑π‘₯β‹―subscript𝑒1π‘₯0u_{dx}=\cdots=u_{1x}=0. Therefore, we have ud,…,u1βˆˆπŠβˆ—subscript𝑒𝑑…subscript𝑒1superscript𝐊u_{d},\ldots,u_{1}\in{\bf K}^{*}. Then equation (3.5)3.5(3.5) has the following form:

u0​x+v0​y+v1​u0​y=0(3.8)subscript𝑒0π‘₯subscript𝑣0𝑦subscript𝑣1subscript𝑒0𝑦03.8u_{0x}+v_{0y}+v_{1}u_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.8)

Since uzβ‰ 0subscript𝑒𝑧0u_{z}\neq 0, so it follows from equation (3.3) that hy​vxβˆ’hx​vy=0subscriptβ„Žπ‘¦subscript𝑣π‘₯subscriptβ„Žπ‘₯subscript𝑣𝑦0h_{y}v_{x}-h_{x}v_{y}=0. That is,

hy​(vm​x​um+β‹―+v1​x​u+v0​x)βˆ’hx​(v(mβˆ’1)​y​umβˆ’1+β‹―+v1​y​u+v0​y)=0(3.9)subscriptβ„Žπ‘¦subscriptπ‘£π‘šπ‘₯superscriptπ‘’π‘šβ‹―subscript𝑣1π‘₯𝑒subscript𝑣0π‘₯subscriptβ„Žπ‘₯subscriptπ‘£π‘š1𝑦superscriptπ‘’π‘š1β‹―subscript𝑣1𝑦𝑒subscript𝑣0𝑦03.9h_{y}(v_{mx}u^{m}+\cdots+v_{1x}u+v_{0x})-h_{x}(v_{(m-1)y}u^{m-1}+\cdots+v_{1y}u+v_{0y})=0~{}~{}~{}~{}~{}(3.9)

Then we have hy​vm​x=0subscriptβ„Žπ‘¦subscriptπ‘£π‘šπ‘₯0h_{y}v_{mx}=0 by comparing the coefficients of zm​dsuperscriptπ‘§π‘šπ‘‘z^{md} of equation (3.9)3.9(3.9). Thus, we have hy=0subscriptβ„Žπ‘¦0h_{y}=0 or vm​x=0subscriptπ‘£π‘šπ‘₯0v_{mx}=0.

(1) If hy=0subscriptβ„Žπ‘¦0h_{y}=0, then it follows from equation (3.9)3.9(3.9) that hx=0subscriptβ„Žπ‘₯0h_{x}=0 or v(mβˆ’1)​y=β‹―=v1​y=v0​y=0subscriptπ‘£π‘š1𝑦⋯subscript𝑣1𝑦subscript𝑣0𝑦0v_{(m-1)y}=\cdots=v_{1y}=v_{0y}=0.

(i) If hx=0subscriptβ„Žπ‘₯0h_{x}=0, then h=0β„Ž0h=0 because h​(0)=0β„Ž00h(0)=0. Thus, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.

(ii) If v(mβˆ’1)​y=β‹―=v0​y=0subscriptπ‘£π‘š1𝑦⋯subscript𝑣0𝑦0v_{(m-1)y}=\cdots=v_{0y}=0, then vy=0subscript𝑣𝑦0v_{y}=0. It follows from equation (3.2)3.2(3.2) that uy​vx+hx​uz=0subscript𝑒𝑦subscript𝑣π‘₯subscriptβ„Žπ‘₯subscript𝑒𝑧0u_{y}v_{x}+h_{x}u_{z}=0. That is, u0​y​(vm​x​um+v(mβˆ’1)​x​umβˆ’1+β‹―+v1​x​u+v0​x)+hx​(d​ud​zdβˆ’1+(dβˆ’1)​udβˆ’1​zdβˆ’2+β‹―+u1)=0subscript𝑒0𝑦subscriptπ‘£π‘šπ‘₯superscriptπ‘’π‘šsubscriptπ‘£π‘š1π‘₯superscriptπ‘’π‘š1β‹―subscript𝑣1π‘₯𝑒subscript𝑣0π‘₯subscriptβ„Žπ‘₯𝑑subscript𝑒𝑑superscript𝑧𝑑1𝑑1subscript𝑒𝑑1superscript𝑧𝑑2β‹―subscript𝑒10u_{0y}(v_{mx}u^{m}+v_{(m-1)x}u^{m-1}+\cdots+v_{1x}u+v_{0x})+h_{x}(du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})=0 Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.10)

If u0​y=0subscript𝑒0𝑦0u_{0y}=0, then it follows from equation (3.10)3.10(3.10) that hx=0subscriptβ„Žπ‘₯0h_{x}=0. Since H​(0)=0𝐻00H(0)=0, we have h=0β„Ž0h=0. Thus, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.

If u0​yβ‰ 0subscript𝑒0𝑦0u_{0y}\neq 0, then we have vm​x=β‹―=v1​x=0subscriptπ‘£π‘šπ‘₯β‹―subscript𝑣1π‘₯0v_{mx}=\cdots=v_{1x}=0 by comparing the coefficients of zm​d,z(mβˆ’1)​d,…,zdsuperscriptπ‘§π‘šπ‘‘superscriptπ‘§π‘š1𝑑…superscript𝑧𝑑z^{md},z^{(m-1)d},\ldots,z^{d} of equation (3.10)3.10(3.10) respectively.
If dβ‰₯2𝑑2d\geq 2, then we have hx=0subscriptβ„Žπ‘₯0h_{x}=0 by comparing the coefficient of zdβˆ’1superscript𝑧𝑑1z^{d-1} of equation (3.10)3.10(3.10). Since H​(0)=0𝐻00H(0)=0, we have h=0β„Ž0h=0. Thus, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.
If d=1𝑑1d=1, then it follows from equation (3.6)3.6(3.6) (j=mβˆ’1π‘—π‘š1j=m-1) that u0​y=0subscript𝑒0𝑦0u_{0y}=0. This is a contradiction!

(2) If vm​x=0subscriptπ‘£π‘šπ‘₯0v_{mx}=0, then vmβˆˆπŠβˆ—subscriptπ‘£π‘šsuperscript𝐊v_{m}\in{\bf K}^{*}. It follows from equation (3.2)3.2(3.2) that
u0​x​(v(mβˆ’1)​y​umβˆ’1+β‹―+v1​y​u+v0​y)βˆ’u0​y​(v(mβˆ’1)​x​umβˆ’1+β‹―+v1​x​u+v0​x)βˆ’hx​(d​ud​zdβˆ’1+(dβˆ’1)​udβˆ’1​zdβˆ’2+β‹―+u1)βˆ’hy​(m​vm​umβˆ’1+β‹―+2​v2​u+v1)​(d​ud​zdβˆ’1+(dβˆ’1)​udβˆ’1​zdβˆ’2+β‹―+u1)=0subscript𝑒0π‘₯subscriptπ‘£π‘š1𝑦superscriptπ‘’π‘š1β‹―subscript𝑣1𝑦𝑒subscript𝑣0𝑦subscript𝑒0𝑦subscriptπ‘£π‘š1π‘₯superscriptπ‘’π‘š1β‹―subscript𝑣1π‘₯𝑒subscript𝑣0π‘₯subscriptβ„Žπ‘₯𝑑subscript𝑒𝑑superscript𝑧𝑑1𝑑1subscript𝑒𝑑1superscript𝑧𝑑2β‹―subscript𝑒1subscriptβ„Žπ‘¦π‘šsubscriptπ‘£π‘šsuperscriptπ‘’π‘š1β‹―2subscript𝑣2𝑒subscript𝑣1𝑑subscript𝑒𝑑superscript𝑧𝑑1𝑑1subscript𝑒𝑑1superscript𝑧𝑑2β‹―subscript𝑒10u_{0x}(v_{(m-1)y}u^{m-1}+\cdots+v_{1y}u+v_{0y})-u_{0y}(v_{(m-1)x}u^{m-1}+\cdots+v_{1x}u+v_{0x})-h_{x}(du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})-h_{y}(mv_{m}u^{m-1}+\cdots+2v_{2}u+v_{1})(du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})=0Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.11)

If dβ‰₯2𝑑2d\geq 2, then we have hy=0subscriptβ„Žπ‘¦0h_{y}=0 by comparing the coefficient of zd​(mβˆ’1)+dβˆ’1superscriptπ‘§π‘‘π‘š1𝑑1z^{d(m-1)+d-1} of equation (3.11)3.11(3.11). Thus, we have u0​x​vl​yβˆ’u0​y​vl​x=0subscript𝑒0π‘₯subscript𝑣𝑙𝑦subscript𝑒0𝑦subscript𝑣𝑙π‘₯0u_{0x}v_{ly}-u_{0y}v_{lx}=0 by comparing the coefficients of zl​dsuperscript𝑧𝑙𝑑z^{ld} of equation (3.11)3.11(3.11) for l=mβˆ’1,mβˆ’2,…,1π‘™π‘š1π‘š2…1l=m-1,m-2,\ldots,1. Comparing the coefficients of zdβˆ’1superscript𝑧𝑑1z^{d-1} of equation (3.11)3.11(3.11), we have hx=0subscriptβ„Žπ‘₯0h_{x}=0. Thus, we have h=0β„Ž0h=0 because h​(0)=0β„Ž00h(0)=0. So u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.

If d=1𝑑1d=1, then equation (3.11)3.11(3.11) has the following form:
u0​x​(v(mβˆ’1)​y​umβˆ’1+β‹―+v1​y​u+v0​y)βˆ’u0​y​(v(mβˆ’1)​x​umβˆ’1+β‹―+v1​x​u+v0​x)βˆ’hx​u1βˆ’hy​u1​(m​vm​umβˆ’1+β‹―+2​v2​u+v1)=0subscript𝑒0π‘₯subscriptπ‘£π‘š1𝑦superscriptπ‘’π‘š1β‹―subscript𝑣1𝑦𝑒subscript𝑣0𝑦subscript𝑒0𝑦subscriptπ‘£π‘š1π‘₯superscriptπ‘’π‘š1β‹―subscript𝑣1π‘₯𝑒subscript𝑣0π‘₯subscriptβ„Žπ‘₯subscript𝑒1subscriptβ„Žπ‘¦subscript𝑒1π‘šsubscriptπ‘£π‘šsuperscriptπ‘’π‘š1β‹―2subscript𝑣2𝑒subscript𝑣10u_{0x}(v_{(m-1)y}u^{m-1}+\cdots+v_{1y}u+v_{0y})-u_{0y}(v_{(m-1)x}u^{m-1}+\cdots+v_{1x}u+v_{0x})-h_{x}u_{1}-h_{y}u_{1}(mv_{m}u^{m-1}+\cdots+2v_{2}u+v_{1})=0Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.12)
Thus, we have

u0​x​vl​yβˆ’u0​y​vl​xβˆ’hy​u1​(l+1)​vl+1=0(3.13)subscript𝑒0π‘₯subscript𝑣𝑙𝑦subscript𝑒0𝑦subscript𝑣𝑙π‘₯subscriptβ„Žπ‘¦subscript𝑒1𝑙1subscript𝑣𝑙103.13u_{0x}v_{ly}-u_{0y}v_{lx}-h_{y}u_{1}(l+1)v_{l+1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.13)

by comparing the coefficients of zl​dsuperscript𝑧𝑙𝑑z^{ld} of equation (3.12)3.12(3.12) for l=mβˆ’1,mβˆ’2,…,1π‘™π‘š1π‘š2…1l=m-1,m-2,\ldots,1. Then equation (3.12)3.12(3.12) has the following form:

u0​x​v0​yβˆ’u0​y​v0​xβˆ’u1​(hx+v1​hy)=0(3.14)subscript𝑒0π‘₯subscript𝑣0𝑦subscript𝑒0𝑦subscript𝑣0π‘₯subscript𝑒1subscriptβ„Žπ‘₯subscript𝑣1subscriptβ„Žπ‘¦03.14u_{0x}v_{0y}-u_{0y}v_{0x}-u_{1}(h_{x}+v_{1}h_{y})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.14)

Comparing the coefficients of zl​dsuperscript𝑧𝑙𝑑z^{ld} of equation (3.9)3.9(3.9) for l=mβˆ’1,mβˆ’2,…,1π‘™π‘š1π‘š2…1l=m-1,m-2,\ldots,1, we have

hy​vl​xβˆ’hx​vl​y=0(3.15)subscriptβ„Žπ‘¦subscript𝑣𝑙π‘₯subscriptβ„Žπ‘₯subscript𝑣𝑙𝑦03.15h_{y}v_{lx}-h_{x}v_{ly}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.15)

for l=mβˆ’1,…,1π‘™π‘š1…1l=m-1,\ldots,1. Then equation (3.9)3.9(3.9) has the following form:

hy​v0​xβˆ’hx​v0​y=0(3.16)subscriptβ„Žπ‘¦subscript𝑣0π‘₯subscriptβ„Žπ‘₯subscript𝑣0𝑦03.16h_{y}v_{0x}-h_{x}v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.16)

It follows from equation (3.6)3.6(3.6) that

vmβˆ’1=βˆ’m​vm​u0+c1​(x)(3.17)subscriptπ‘£π‘š1π‘šsubscriptπ‘£π‘šsubscript𝑒0subscript𝑐1π‘₯3.17v_{m-1}=-mv_{m}u_{0}+c_{1}(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.17)

for some c1​(x)βˆˆπŠβ€‹[x]subscript𝑐1π‘₯𝐊delimited-[]π‘₯c_{1}(x)\in{\bf K}[x]. Substituting equation (3.17) to equation (3.13) for l=mβˆ’1π‘™π‘š1l=m-1, we have

u0​y​c1′​(x)+hy​m​vm​u1=0subscript𝑒0𝑦subscriptsuperscript𝑐′1π‘₯subscriptβ„Žπ‘¦π‘šsubscriptπ‘£π‘šsubscript𝑒10u_{0y}c^{\prime}_{1}(x)+h_{y}mv_{m}u_{1}=0

Thus, we have u0​c1′​(x)+m​vm​u1β‹…h+e​(x)=0subscript𝑒0subscriptsuperscript𝑐′1π‘₯β‹…π‘šsubscriptπ‘£π‘šsubscript𝑒1β„Žπ‘’π‘₯0u_{0}c^{\prime}_{1}(x)+mv_{m}u_{1}\cdot h+e(x)=0 by integrating the two sides of the above equation with respect to y𝑦y, where e​(x)βˆˆπŠβ€‹[x]𝑒π‘₯𝐊delimited-[]π‘₯e(x)\in{\bf K}[x]. That is,

h=βˆ’1m​vm​u1​c1′​(x)​u0βˆ’1m​vm​u1​e​(x)(3.18)β„Ž1π‘šsubscriptπ‘£π‘šsubscript𝑒1subscriptsuperscript𝑐′1π‘₯subscript𝑒01π‘šsubscriptπ‘£π‘šsubscript𝑒1𝑒π‘₯3.18h=-\frac{1}{mv_{m}u_{1}}c^{\prime}_{1}(x)u_{0}-\frac{1}{mv_{m}u_{1}}e(x)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.18)

Substituting equations (3.17)3.17(3.17) and (3.18)3.18(3.18) to equation (3.15)3.15(3.15) for l=mβˆ’1π‘™π‘š1l=m-1, we have

c1′′​(x)​u0​u0​y+1m​vm​(c1′​(x))2​u0​y+e′​(x)​u0​y=0subscriptsuperscript𝑐′′1π‘₯subscript𝑒0subscript𝑒0𝑦1π‘šsubscriptπ‘£π‘šsuperscriptsubscriptsuperscript𝑐′1π‘₯2subscript𝑒0𝑦superscript𝑒′π‘₯subscript𝑒0𝑦0c^{\prime\prime}_{1}(x)u_{0}u_{0y}+\frac{1}{mv_{m}}(c^{\prime}_{1}(x))^{2}u_{0y}+e^{\prime}(x)u_{0y}=0

If u0​y=0subscript𝑒0𝑦0u_{0y}=0, then it follows from equation (3.18)3.18(3.18) that hy=0subscriptβ„Žπ‘¦0h_{y}=0. Then it reduces to (1).
If u0​yβ‰ 0subscript𝑒0𝑦0u_{0y}\neq 0, then we have

c1′′​(x)​u0+1m​vm​(c1′​(x))2+e′​(x)=0(3.19)subscriptsuperscript𝑐′′1π‘₯subscript𝑒01π‘šsubscriptπ‘£π‘šsuperscriptsubscriptsuperscript𝑐′1π‘₯2superscript𝑒′π‘₯03.19c^{\prime\prime}_{1}(x)u_{0}+\frac{1}{mv_{m}}(c^{\prime}_{1}(x))^{2}+e^{\prime}(x)=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.19)

Thus, we have c1′′​(x)=0superscriptsubscript𝑐1β€²β€²π‘₯0c_{1}^{\prime\prime}(x)=0 and

e′​(x)=βˆ’1m​vm​(c1′​(x))2∈𝐊(3.20)formulae-sequencesuperscript𝑒′π‘₯1π‘šsubscriptπ‘£π‘šsuperscriptsuperscriptsubscript𝑐1β€²π‘₯2𝐊3.20e^{\prime}(x)=-\frac{1}{mv_{m}}(c_{1}^{\prime}(x))^{2}\in{\bf K}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.20)

by comparing the degree of y𝑦y of the equation (3.19). Then we have

c1′​(x)​(u0​x​v0​yβˆ’u0​y​v0​x)+e′​(x)​v0​y=0(3.21)superscriptsubscript𝑐1β€²π‘₯subscript𝑒0π‘₯subscript𝑣0𝑦subscript𝑒0𝑦subscript𝑣0π‘₯superscript𝑒′π‘₯subscript𝑣0𝑦03.21c_{1}^{\prime}(x)(u_{0x}v_{0y}-u_{0y}v_{0x})+e^{\prime}(x)v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.21)

by substituting equation (3.18)3.18(3.18) to equation (3.16)3.16(3.16). Substituting equation (3.20) to equation (3.21), we have

c1′​(x)​(u0​x​v0​yβˆ’u0​y​v0​x)βˆ’1m​vm​(c1′​(x))2​v0​y=0(3.22)superscriptsubscript𝑐1β€²π‘₯subscript𝑒0π‘₯subscript𝑣0𝑦subscript𝑒0𝑦subscript𝑣0π‘₯1π‘šsubscriptπ‘£π‘šsuperscriptsuperscriptsubscript𝑐1β€²π‘₯2subscript𝑣0𝑦03.22c_{1}^{\prime}(x)(u_{0x}v_{0y}-u_{0y}v_{0x})-\frac{1}{mv_{m}}(c_{1}^{\prime}(x))^{2}v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.22)

If c1′​(x)=0superscriptsubscript𝑐1β€²π‘₯0c_{1}^{\prime}(x)=0, then it follows from equation (3.20) that e​(x)βˆˆπŠπ‘’π‘₯𝐊e(x)\in{\bf K}. Thus, it follows from equation (3.18) that h=βˆ’e​(x)m​vm​u1βˆˆπŠβ„Žπ‘’π‘₯π‘šsubscriptπ‘£π‘šsubscript𝑒1𝐊h=-\frac{e(x)}{mv_{m}u_{1}}\in{\bf K}. Since h​(0,0)=0β„Ž000h(0,0)=0, we have h=0β„Ž0h=0. Therefore, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.

If c1′​(x)β‰ 0superscriptsubscript𝑐1β€²π‘₯0c_{1}^{\prime}(x)\neq 0, then equation (3.22) has the following form:

u0​x​v0​yβˆ’u0​y​v0​x=1m​vm​c1′​(x)​v0​y(3.23)subscript𝑒0π‘₯subscript𝑣0𝑦subscript𝑒0𝑦subscript𝑣0π‘₯1π‘šsubscriptπ‘£π‘šsuperscriptsubscript𝑐1β€²π‘₯subscript𝑣0𝑦3.23u_{0x}v_{0y}-u_{0y}v_{0x}=\frac{1}{mv_{m}}c_{1}^{\prime}(x)v_{0y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.23)

Substituting equations (3.18) and (3.23) to equation (3.14), we have the following equation:

c1′​(x)​(v0​y+u0​x+v1​u0​y)+e′​(x)=0.superscriptsubscript𝑐1β€²π‘₯subscript𝑣0𝑦subscript𝑒0π‘₯subscript𝑣1subscript𝑒0𝑦superscript𝑒′π‘₯0c_{1}^{\prime}(x)(v_{0y}+u_{0x}+v_{1}u_{0y})+e^{\prime}(x)=0.

Substituting equation (3.8) to the above equation, we have e′​(x)=0superscript𝑒′π‘₯0e^{\prime}(x)=0. It follows from equation (3.20) that c1′​(x)=0superscriptsubscript𝑐1β€²π‘₯0c_{1}^{\prime}(x)=0. This is a contradiction! ∎

Theorem 3.3.

Let H=(u​(x,y,z),v​(x,y,u),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦π‘’β„Žπ‘₯𝑦H=(u(x,y,z),v(x,y,u),h(x,y)) be a polynomial map over πŠβ€‹[x,y,z]𝐊π‘₯𝑦𝑧{\bf K}[x,y,z]. Assume that H​(0)=0𝐻00H(0)=0 and the components of H𝐻H are linearly independent over 𝐊𝐊{\bf K}. If J​H𝐽𝐻JH is nilpotent and (deg⁑v​(x,y,0),deg⁑h)≀3degree𝑣π‘₯𝑦0degreeβ„Ž3(\deg v(x,y,0),\deg h)\leq 3, then there exists T∈GL3⁑(𝐊)𝑇subscriptGL3𝐊T\in\operatorname{GL}_{3}({\bf K}) such that Tβˆ’1​H​Tsuperscript𝑇1𝐻𝑇T^{-1}HT has the form of Proposition 3.1.

Proof.

It follows from Theorem 3.2 that degu⁑v≀1subscriptdegree𝑒𝑣1\deg_{u}v\leq 1. If uz=0subscript𝑒𝑧0u_{z}=0 or vu=0subscript𝑣𝑒0v_{u}=0, then it reduces to Proposition 2.1 in [12] and Proposition 3.1 respectively. Thus, we can assume that uzβ‰ 0subscript𝑒𝑧0u_{z}\neq 0 and vuβ‰ 0subscript𝑣𝑒0v_{u}\neq 0. Since J​H𝐽𝐻JH is nilpotent, we have the following equations:

{ux+vy=βˆ’vu​uy(3.1)ux​vyβˆ’vx​uyβˆ’hx​uz=hy​vu​uz(3.2)uz​(vx​hyβˆ’vy​hx)=0(3.3)\left\{\begin{aligned} u_{x}+v_{y}=-v_{u}u_{y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.1)\\ u_{x}v_{y}-v_{x}u_{y}-h_{x}u_{z}=h_{y}v_{u}u_{z}~{}~{}~{}~{}~{}~{}~{}~{}(3.2)\\ u_{z}(v_{x}h_{y}-v_{y}h_{x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.3)\\ \end{aligned}\right.

Let u=ud​zd+udβˆ’1​zdβˆ’1+β‹―+u1​z+u0𝑒subscript𝑒𝑑superscript𝑧𝑑subscript𝑒𝑑1superscript𝑧𝑑1β‹―subscript𝑒1𝑧subscript𝑒0u=u_{d}z^{d}+u_{d-1}z^{d-1}+\cdots+u_{1}z+u_{0} and v=v1​u+v0𝑣subscript𝑣1𝑒subscript𝑣0v=v_{1}u+v_{0} with ud​v1β‰ 0subscript𝑒𝑑subscript𝑣10u_{d}v_{1}\neq 0.

It follows from equation (3.1) that ud​x​zd+u(dβˆ’1)​x​zdβˆ’1+β‹―+u1​x​z+u0​x+v1​y​u+v0​y+v1​(ud​y​zd+u(dβˆ’1)​y​zdβˆ’1+β‹―+u1​y​z+u0​y)=0subscript𝑒𝑑π‘₯superscript𝑧𝑑subscript𝑒𝑑1π‘₯superscript𝑧𝑑1β‹―subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣1𝑦𝑒subscript𝑣0𝑦subscript𝑣1subscript𝑒𝑑𝑦superscript𝑧𝑑subscript𝑒𝑑1𝑦superscript𝑧𝑑1β‹―subscript𝑒1𝑦𝑧subscript𝑒0𝑦0u_{dx}z^{d}+u_{(d-1)x}z^{d-1}+\cdots+u_{1x}z+u_{0x}+v_{1y}u+v_{0y}+v_{1}(u_{dy}z^{d}+u_{(d-1)y}z^{d-1}+\cdots+u_{1y}z+u_{0y})=0 Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.24)
Then we have

ud​x+v1​y​ud+v1​ud​y=0(3.25)subscript𝑒𝑑π‘₯subscript𝑣1𝑦subscript𝑒𝑑subscript𝑣1subscript𝑒𝑑𝑦03.25u_{dx}+v_{1y}u_{d}+v_{1}u_{dy}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.25)

and

u(dβˆ’1)​x+v1​y​udβˆ’1+v1​u(dβˆ’1)​y=0(3.26)subscript𝑒𝑑1π‘₯subscript𝑣1𝑦subscript𝑒𝑑1subscript𝑣1subscript𝑒𝑑1𝑦03.26u_{(d-1)x}+v_{1y}u_{d-1}+v_{1}u_{(d-1)y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.26)

by comparing the coefficients of zdsuperscript𝑧𝑑z^{d} and zdβˆ’1superscript𝑧𝑑1z^{d-1} of equation (3.24) respectively.

It follows from equation (3.2) that
(ud​x​zd+u(dβˆ’1)​x​zdβˆ’1+β‹―+u1​x​z+u0​x)​(v1​y​u+v0​y)βˆ’(ud​y​zd+u(dβˆ’1)​y​zdβˆ’1+β‹―+u1​y​z+u0​y)​(v1​x​u+v0​x)βˆ’hx​(d​ud​zdβˆ’1+(dβˆ’1)​udβˆ’1​zdβˆ’2+β‹―+u1)βˆ’hy​v1​(d​ud​zdβˆ’1+(dβˆ’1)​udβˆ’1​zdβˆ’2+β‹―+u1)=0subscript𝑒𝑑π‘₯superscript𝑧𝑑subscript𝑒𝑑1π‘₯superscript𝑧𝑑1β‹―subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣1𝑦𝑒subscript𝑣0𝑦subscript𝑒𝑑𝑦superscript𝑧𝑑subscript𝑒𝑑1𝑦superscript𝑧𝑑1β‹―subscript𝑒1𝑦𝑧subscript𝑒0𝑦subscript𝑣1π‘₯𝑒subscript𝑣0π‘₯subscriptβ„Žπ‘₯𝑑subscript𝑒𝑑superscript𝑧𝑑1𝑑1subscript𝑒𝑑1superscript𝑧𝑑2β‹―subscript𝑒1subscriptβ„Žπ‘¦subscript𝑣1𝑑subscript𝑒𝑑superscript𝑧𝑑1𝑑1subscript𝑒𝑑1superscript𝑧𝑑2β‹―subscript𝑒10(u_{dx}z^{d}+u_{(d-1)x}z^{d-1}+\cdots+u_{1x}z+u_{0x})(v_{1y}u+v_{0y})-(u_{dy}z^{d}+u_{(d-1)y}z^{d-1}+\cdots+u_{1y}z+u_{0y})(v_{1x}u+v_{0x})-h_{x}(du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})-h_{y}v_{1}(du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})=0Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (3.27)
Comparing the coefficients of zksuperscriptπ‘§π‘˜z^{k} of equation (3.27) for k=2​d,2​dβˆ’1,…,d+1π‘˜2𝑑2𝑑1…𝑑1k=2d,2d-1,\ldots,d+1, we have

u(kβˆ’d)​x​v1​yβˆ’u(kβˆ’d)​y​v1​x=0(3.28)subscriptπ‘’π‘˜π‘‘π‘₯subscript𝑣1𝑦subscriptπ‘’π‘˜π‘‘π‘¦subscript𝑣1π‘₯03.28u_{(k-d)x}v_{1y}-u_{(k-d)y}v_{1x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.28)

Then we have

(u0​x​v1​yβˆ’u0​y​v1​x)​ud+ud​x​v0​yβˆ’ud​y​v0​x=0(3.29)subscript𝑒0π‘₯subscript𝑣1𝑦subscript𝑒0𝑦subscript𝑣1π‘₯subscript𝑒𝑑subscript𝑒𝑑π‘₯subscript𝑣0𝑦subscript𝑒𝑑𝑦subscript𝑣0π‘₯03.29(u_{0x}v_{1y}-u_{0y}v_{1x})u_{d}+u_{dx}v_{0y}-u_{dy}v_{0x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.29)

and

(u0​x​v1​yβˆ’u0​y​v1​x)​udβˆ’1+u(dβˆ’1)​x​v0​yβˆ’u(dβˆ’1)​y​v0​xβˆ’d​ud​(hx+v1​hy)=0(3.30)subscript𝑒0π‘₯subscript𝑣1𝑦subscript𝑒0𝑦subscript𝑣1π‘₯subscript𝑒𝑑1subscript𝑒𝑑1π‘₯subscript𝑣0𝑦subscript𝑒𝑑1𝑦subscript𝑣0π‘₯𝑑subscript𝑒𝑑subscriptβ„Žπ‘₯subscript𝑣1subscriptβ„Žπ‘¦03.30(u_{0x}v_{1y}-u_{0y}v_{1x})u_{d-1}+u_{(d-1)x}v_{0y}-u_{(d-1)y}v_{0x}-du_{d}(h_{x}+v_{1}h_{y})=0~{}~{}~{}~{}~{}~{}(3.30)

by comparing the coefficients of zdsuperscript𝑧𝑑z^{d} and zdβˆ’1superscript𝑧𝑑1z^{d-1} of equation (3.27) respectively.

It follows from equation (3.3) that

hy​(v1​x​u+v0​x)βˆ’hx​(v1​y​u+v0​y)=0(3.31)subscriptβ„Žπ‘¦subscript𝑣1π‘₯𝑒subscript𝑣0π‘₯subscriptβ„Žπ‘₯subscript𝑣1𝑦𝑒subscript𝑣0𝑦03.31h_{y}(v_{1x}u+v_{0x})-h_{x}(v_{1y}u+v_{0y})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.31)

Then we have

hy​v1​xβˆ’hx​v1​y=0(3.32)subscriptβ„Žπ‘¦subscript𝑣1π‘₯subscriptβ„Žπ‘₯subscript𝑣1𝑦03.32h_{y}v_{1x}-h_{x}v_{1y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.32)

and

hy​v0​xβˆ’hx​v0​y=0(3.33)subscriptβ„Žπ‘¦subscript𝑣0π‘₯subscriptβ„Žπ‘₯subscript𝑣0𝑦03.33h_{y}v_{0x}-h_{x}v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.33)

by comparing the coefficients of zdsuperscript𝑧𝑑z^{d} and z0superscript𝑧0z^{0} of equation (3.31) orderly.

If hyβ‰ 0subscriptβ„Žπ‘¦0h_{y}\neq 0, then let λ​(x,y)=hxhyπœ†π‘₯𝑦subscriptβ„Žπ‘₯subscriptβ„Žπ‘¦\lambda(x,y)=\frac{h_{x}}{h_{y}}. Thus, we have hx=λ​(x,y)​hysubscriptβ„Žπ‘₯πœ†π‘₯𝑦subscriptβ„Žπ‘¦h_{x}=\lambda(x,y)h_{y}. It follows from equations (3.32) and (3.33) that

v1​x=λ​(x,y)​v1​y(3.34)subscript𝑣1π‘₯πœ†π‘₯𝑦subscript𝑣1𝑦3.34v_{1x}=\lambda(x,y)v_{1y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.34)

and

v0​x=λ​(x,y)​v0​y(3.35)subscript𝑣0π‘₯πœ†π‘₯𝑦subscript𝑣0𝑦3.35v_{0x}=\lambda(x,y)v_{0y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.35)

It follows from equations (3.28) and (3.34) that

(u(kβˆ’d)​xβˆ’Ξ»β€‹(x,y)​u(kβˆ’d)​y)​v1​y=0subscriptπ‘’π‘˜π‘‘π‘₯πœ†π‘₯𝑦subscriptπ‘’π‘˜π‘‘π‘¦subscript𝑣1𝑦0(u_{(k-d)x}-\lambda(x,y)u_{(k-d)y})v_{1y}=0

If v1​y=0subscript𝑣1𝑦0v_{1y}=0, then it follows from equation (3.32) that v1​x=0subscript𝑣1π‘₯0v_{1x}=0 because hyβ‰ 0subscriptβ„Žπ‘¦0h_{y}\neq 0. That is, v1βˆˆπŠβˆ—subscript𝑣1superscript𝐊v_{1}\in{\bf K}^{*}.

If v1​yβ‰ 0subscript𝑣1𝑦0v_{1y}\neq 0, then we have

u(kβˆ’d)​x=λ​(x,y)​u(kβˆ’d)​y(3.36)subscriptπ‘’π‘˜π‘‘π‘₯πœ†π‘₯𝑦subscriptπ‘’π‘˜π‘‘π‘¦3.36u_{(k-d)x}=\lambda(x,y)u_{(k-d)y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.36)

for d+1≀k≀2​d𝑑1π‘˜2𝑑d+1\leq k\leq 2d. Substituting equations (3.34), (3.35), (3.36)(k=2​d)π‘˜2𝑑(k=2d) to equation (3.29), we have the following equation:

u0​x=λ​(x,y)​u0​y(3.37)subscript𝑒0π‘₯πœ†π‘₯𝑦subscript𝑒0𝑦3.37u_{0x}=\lambda(x,y)u_{0y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3.37)

Substituting equations (3.34), (3.35), (3.36), (3.37) to equation (3.30), we have the following equation:

hx+v1​hy=0.subscriptβ„Žπ‘₯subscript𝑣1subscriptβ„Žπ‘¦0h_{x}+v_{1}h_{y}=0.

That is,

v1=βˆ’hxhy=βˆ’Ξ»β€‹(x,y).subscript𝑣1subscriptβ„Žπ‘₯subscriptβ„Žπ‘¦πœ†π‘₯𝑦v_{1}=-\frac{h_{x}}{h_{y}}=-\lambda(x,y).

Substituting equation (3.34) to the above equation, we have

v1​v1​y=βˆ’v1​x.subscript𝑣1subscript𝑣1𝑦subscript𝑣1π‘₯v_{1}v_{1y}=-v_{1x}.

Then we have v1βˆˆπŠβˆ—subscript𝑣1superscript𝐊v_{1}\in{\bf K}^{*} by comparing the degree of xπ‘₯x of two sides of the above equation. This is a contradiction!
Consequently, we have H=(u,v1​u+v0​(x,y),h​(x,y))𝐻𝑒subscript𝑣1𝑒subscript𝑣0π‘₯π‘¦β„Žπ‘₯𝑦H=(u,v_{1}u+v_{0}(x,y),h(x,y)) with v1βˆˆπŠβˆ—subscript𝑣1superscript𝐊v_{1}\in{\bf K}^{*}. Let

T=(100v110001).𝑇100subscript𝑣110001T=\left(\begin{array}[]{ccc}1&0&0\\ v_{1}&1&0\\ 0&0&1\\ \end{array}\right).

Then Tβˆ’1​H​T=(u​(x,y+v1​x,z),v0​(x,y+v1​x),h​(x,y+v1​x))superscript𝑇1𝐻𝑇𝑒π‘₯𝑦subscript𝑣1π‘₯𝑧subscript𝑣0π‘₯𝑦subscript𝑣1π‘₯β„Žπ‘₯𝑦subscript𝑣1π‘₯T^{-1}HT=(u(x,y+v_{1}x,z),v_{0}(x,y+v_{1}x),h(x,y+v_{1}x)) and degx⁑v0​(x,y+v1​x)=deg⁑v​(x,y,0)subscriptdegreeπ‘₯subscript𝑣0π‘₯𝑦subscript𝑣1π‘₯degree𝑣π‘₯𝑦0\deg_{x}v_{0}(x,y+v_{1}x)=\deg v(x,y,0), degx⁑h​(x,y+v1​x)=deg⁑h​(x,y)subscriptdegreeπ‘₯β„Žπ‘₯𝑦subscript𝑣1π‘₯degreeβ„Žπ‘₯𝑦\deg_{x}h(x,y+v_{1}x)=\deg h(x,y). Since (deg⁑v​(x,y,0),deg⁑h)≀3degree𝑣π‘₯𝑦0degreeβ„Ž3(\deg v(x,y,0),\deg h)\leq 3, so the conclusion follows from Proposition 3.1.

If hy=0subscriptβ„Žπ‘¦0h_{y}=0, then it follows from equation (3.32) that hx=0subscriptβ„Žπ‘₯0h_{x}=0 or v1​y=0subscript𝑣1𝑦0v_{1y}=0.

(1) If hx=0subscriptβ„Žπ‘₯0h_{x}=0, then h=0β„Ž0h=0 because h​(0,0)=0β„Ž000h(0,0)=0. Thus, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent. This is a contradiction!

(2) If v1​y=0subscript𝑣1𝑦0v_{1y}=0, then it follows from equation (3.28)3.28(3.28) that v1​x=0subscript𝑣1π‘₯0v_{1x}=0 or ud​y=u(dβˆ’1)​y=β‹―=u1​y=0subscript𝑒𝑑𝑦subscript𝑒𝑑1𝑦⋯subscript𝑒1𝑦0u_{dy}=u_{(d-1)y}=\cdots=u_{1y}=0.

Case I If v1​x=0subscript𝑣1π‘₯0v_{1x}=0, then v1βˆˆπŠβˆ—subscript𝑣1superscript𝐊v_{1}\in{\bf K}^{*}. Thus, the conclusion follows from the former arguments.

Case II If ud​y=u(dβˆ’1)​y=β‹―=u1​y=0subscript𝑒𝑑𝑦subscript𝑒𝑑1𝑦⋯subscript𝑒1𝑦0u_{dy}=u_{(d-1)y}=\cdots=u_{1y}=0, then it follows from equations (3.25) and (3.26) that ud​x=0=u(dβˆ’1)​xsubscript𝑒𝑑π‘₯0subscript𝑒𝑑1π‘₯u_{dx}=0=u_{(d-1)x}. Thus, it follows from equation (3.29) that v1​x=0subscript𝑣1π‘₯0v_{1x}=0 or u0​y=0subscript𝑒0𝑦0u_{0y}=0. If v1​x=0subscript𝑣1π‘₯0v_{1x}=0, then we have v1βˆˆπŠβˆ—subscript𝑣1superscript𝐊v_{1}\in{\bf K}^{*}, it reduces to Case I. If u0​y=0subscript𝑒0𝑦0u_{0y}=0, then it follows from equation (3.30) that hx=0subscriptβ„Žπ‘₯0h_{x}=0. Thus, we have h=0β„Ž0h=0 because h​(0,0)=0β„Ž000h(0,0)=0. Therefore, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent. This is a contradiction! ∎

Corollary 3.4.

Let H=(u​(x,y,v),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑣𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y,v),v(x,y,z),h(x,y)) be a polynomial map with H​(0)=0𝐻00H(0)=0. Assume that the components of H𝐻H are linearly independent over 𝐊𝐊{\bf K}. If J​H𝐽𝐻JH is nilpotent and (deg⁑u​(x,y,0),deg⁑h​(x,y))≀3degree𝑒π‘₯𝑦0degreeβ„Žπ‘₯𝑦3(\deg u(x,y,0),\deg h(x,y))\leq 3, then there exists T∈GL3⁑(𝐊)𝑇subscriptGL3𝐊T\in\operatorname{GL}_{3}({\bf K}) such that Tβˆ’1​H​Tsuperscript𝑇1𝐻𝑇T^{-1}HT has the form of Proposition 3.1.

Proof.

Let

T1=(010100001).subscript𝑇1010100001T_{1}=\left(\begin{array}[]{ccc}0&1&0\\ 1&0&0\\ 0&0&1\\ \end{array}\right).

Then T1βˆ’1​H​T1=(v​(y,x,z),u​(y,x,v​(y,x,z)),h​(y,x))superscriptsubscript𝑇11𝐻subscript𝑇1𝑣𝑦π‘₯𝑧𝑒𝑦π‘₯𝑣𝑦π‘₯π‘§β„Žπ‘¦π‘₯T_{1}^{-1}HT_{1}=(v(y,x,z),u(y,x,v(y,x,z)),h(y,x)). Since J​H𝐽𝐻JH is nilpotent, we have that J​(T1βˆ’1​H​T1)=T1βˆ’1​J​H​T1𝐽superscriptsubscript𝑇11𝐻subscript𝑇1superscriptsubscript𝑇11𝐽𝐻subscript𝑇1J(T_{1}^{-1}HT_{1})=T_{1}^{-1}JHT_{1} is nilpotent. Since deg⁑u​(y,x,0)=deg⁑u​(x,y,0)degree𝑒𝑦π‘₯0degree𝑒π‘₯𝑦0\deg u(y,x,0)=\deg u(x,y,0), degh(y,x))=degh(x,y)\deg h(y,x))=\deg h(x,y) and (deg⁑u​(x,y,0),deg⁑h​(x,y))≀3degree𝑒π‘₯𝑦0degreeβ„Žπ‘₯𝑦3(\deg u(x,y,0),\deg h(x,y))\leq 3, so it follows from Theorem 3.3 that there exists T2∈GL3⁑(𝐊)subscript𝑇2subscriptGL3𝐊T_{2}\in\operatorname{GL}_{3}({\bf K}) such that T2βˆ’1​(T1βˆ’1​H​T1)​T2superscriptsubscript𝑇21superscriptsubscript𝑇11𝐻subscript𝑇1subscript𝑇2T_{2}^{-1}(T_{1}^{-1}HT_{1})T_{2} is of the form of Proposition 3.1. Let T=T1​T2∈GL3⁑(𝐊)𝑇subscript𝑇1subscript𝑇2subscriptGL3𝐊T=T_{1}T_{2}\in\operatorname{GL}_{3}({\bf K}). Then the conclusion follows. ∎

4 Polynomial maps of the form H=(u​(x,y,z),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y,z),v(x,\allowbreak y,z),h(x,y))

In the section, we classify polynomial maps of the form H=(u​(x,y,z),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y,z),v(x,y,z),\allowbreak h(x,y)) in the case that J​H𝐽𝐻JH is nilpotent and degz⁑v​(x,y,z)≀1subscriptdegree𝑧𝑣π‘₯𝑦𝑧1\deg_{z}v(x,y,z)\leq 1. Firstly, we prove that u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent in the case that J​H𝐽𝐻JH is nilpotent and degz⁑v​(x,y,z)=1subscriptdegree𝑧𝑣π‘₯𝑦𝑧1\deg_{z}v(x,y,z)=1 and degz⁑uβ‰₯2subscriptdegree𝑧𝑒2\deg_{z}u\geq 2. In the proof of the following theorem, we divide two cases according to the degree of z𝑧z in u𝑒u. Case I: degz⁑u=2subscriptdegree𝑧𝑒2\deg_{z}u=2; Case II: degz⁑uβ‰₯3subscriptdegree𝑧𝑒3\deg_{z}u\geq 3. In the first case, we divide two parts. We have three subcases in the second case.

Theorem 4.1.

Let H=(u​(x,y,z),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y,z),v(x,y,z),h(x,y)) be a polynomial map with degz⁑v​(x,y,z)=1subscriptdegree𝑧𝑣π‘₯𝑦𝑧1\deg_{z}v(x,y,z)=1. Assume that H​(0)=0𝐻00H(0)=0 and degz⁑uβ‰₯2subscriptdegree𝑧𝑒2\deg_{z}u\geq 2. If J​H𝐽𝐻JH is nilpotent, then u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.

Proof.

Let u=ud​zd+β‹―+u1​z+u0𝑒subscript𝑒𝑑superscript𝑧𝑑⋯subscript𝑒1𝑧subscript𝑒0u=u_{d}z^{d}+\cdots+u_{1}z+u_{0}, v=v1​z+v0𝑣subscript𝑣1𝑧subscript𝑣0v=v_{1}z+v_{0} with ud​v1β‰ 0subscript𝑒𝑑subscript𝑣10u_{d}v_{1}\neq 0 and dβ‰₯2𝑑2d\geq 2. Since J​H𝐽𝐻JH is nilpotent, we have the following equations:

{ux+vy=0(4.1)ux​vyβˆ’vx​uyβˆ’hx​uzβˆ’hy​vz=0(4.2)vx​hy​uzβˆ’hx​vy​uz+hx​uy​vzβˆ’ux​hy​vz=0(4.3)\left\{\begin{aligned} u_{x}+v_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.1)\\ u_{x}v_{y}-v_{x}u_{y}-h_{x}u_{z}-h_{y}v_{z}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.2)\\ v_{x}h_{y}u_{z}-h_{x}v_{y}u_{z}+h_{x}u_{y}v_{z}-u_{x}h_{y}v_{z}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.3)\\ \end{aligned}\right.

It follows from equation (4.1) that

ud​x​zd+u(dβˆ’1)​x​zdβˆ’1+β‹―+u1​x​z+u0​x+v1​y​z+v0​y=0(4.4)subscript𝑒𝑑π‘₯superscript𝑧𝑑subscript𝑒𝑑1π‘₯superscript𝑧𝑑1β‹―subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣1𝑦𝑧subscript𝑣0𝑦04.4u_{dx}z^{d}+u_{(d-1)x}z^{d-1}+\cdots+u_{1x}z+u_{0x}+v_{1y}z+v_{0y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.4)

We always view that the polynomials are in πŠβ€‹[x,y]​[z]𝐊π‘₯𝑦delimited-[]𝑧{\bf K}[x,y][z] with coefficients in πŠβ€‹[x,y]​[z]𝐊π‘₯𝑦delimited-[]𝑧{\bf K}[x,y][z] in the following arguments. Thus, we have

ud​x=β‹―=u2​x=0(4.5)formulae-sequencesubscript𝑒𝑑π‘₯β‹―subscript𝑒2π‘₯04.5u_{dx}=\cdots=u_{2x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.5)

and

ui​x+vi​y=0(4.6)subscript𝑒𝑖π‘₯subscript𝑣𝑖𝑦04.6u_{ix}+v_{iy}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.6)

for i=1,0𝑖10i=1,~{}0 by comparing the coefficients of zjsuperscript𝑧𝑗z^{j} of equation (4.4)4.4(4.4) for j=d,…,1,0𝑗𝑑…10j=d,\ldots,1,0.

It follows from equations (4.2) and (4.5) that
(u1​x​z+u0​x)​(v1​y​z+v0​y)βˆ’(v1​x​z+v0​x)​(ud​y​zd+u(dβˆ’1)​y​zdβˆ’1+β‹―+u1​y​z+u0​y)βˆ’hx​(d​ud​zdβˆ’1+β‹―+u1)βˆ’hy​v1=0subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣1𝑦𝑧subscript𝑣0𝑦subscript𝑣1π‘₯𝑧subscript𝑣0π‘₯subscript𝑒𝑑𝑦superscript𝑧𝑑subscript𝑒𝑑1𝑦superscript𝑧𝑑1β‹―subscript𝑒1𝑦𝑧subscript𝑒0𝑦subscriptβ„Žπ‘₯𝑑subscript𝑒𝑑superscript𝑧𝑑1β‹―subscript𝑒1subscriptβ„Žπ‘¦subscript𝑣10(u_{1x}z+u_{0x})(v_{1y}z+v_{0y})-(v_{1x}z+v_{0x})(u_{dy}z^{d}+u_{(d-1)y}z^{d-1}+\cdots+u_{1y}z+u_{0y})-h_{x}(du_{d}z^{d-1}+\cdots+u_{1})-h_{y}v_{1}=0 Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (4.7)

Case I If d=2𝑑2d=2, then comparing the coefficients of z3superscript𝑧3z^{3} and z2superscript𝑧2z^{2} of equation (4.7), we have v1​x​u2​y=0subscript𝑣1π‘₯subscript𝑒2𝑦0v_{1x}u_{2y}=0 and

u1​x​v1​yβˆ’v1​x​u1​yβˆ’v0​x​u2​y=0(4.8)subscript𝑒1π‘₯subscript𝑣1𝑦subscript𝑣1π‘₯subscript𝑒1𝑦subscript𝑣0π‘₯subscript𝑒2𝑦04.8u_{1x}v_{1y}-v_{1x}u_{1y}-v_{0x}u_{2y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.8)

Thus, we have u2​y=0subscript𝑒2𝑦0u_{2y}=0 or v1​x=0subscript𝑣1π‘₯0v_{1x}=0.

(1) If u2​y=0subscript𝑒2𝑦0u_{2y}=0, then it follows from equations (4.5) and (4.8) that u2βˆˆπŠβˆ—subscript𝑒2superscript𝐊u_{2}\in{\bf K}^{*} and

u1​x​v1​yβˆ’v1​x​u1​y=0(4.9)subscript𝑒1π‘₯subscript𝑣1𝑦subscript𝑣1π‘₯subscript𝑒1𝑦04.9u_{1x}v_{1y}-v_{1x}u_{1y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.9)

It follows from equation (4.6) that u1​x=βˆ’v1​ysubscript𝑒1π‘₯subscript𝑣1𝑦u_{1x}=-v_{1y}, so there exists PβˆˆπŠβ€‹[x,y]π‘ƒπŠπ‘₯𝑦P\in{\bf K}[x,y] such that u1=Pysubscript𝑒1subscript𝑃𝑦u_{1}=P_{y}, v1=βˆ’Pxsubscript𝑣1subscript𝑃π‘₯v_{1}=-P_{x}. It follows from equation (4.9) that Px​y2βˆ’Px​x​Py​y=0superscriptsubscript𝑃π‘₯𝑦2subscript𝑃π‘₯π‘₯subscript𝑃𝑦𝑦0P_{xy}^{2}-P_{xx}P_{yy}=0. Then it follows from Lemma 2.1 in [4] that

u1=Py=b​f​(a​x+b​y)+c2(4.10)formulae-sequencesubscript𝑒1subscriptπ‘ƒπ‘¦π‘π‘“π‘Žπ‘₯𝑏𝑦subscript𝑐24.10u_{1}=P_{y}=bf(ax+by)+c_{2}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.10)

and

v1=βˆ’Px=βˆ’a​f​(a​x+b​y)+c1(4.11)formulae-sequencesubscript𝑣1subscript𝑃π‘₯π‘Žπ‘“π‘Žπ‘₯𝑏𝑦subscript𝑐14.11v_{1}=-P_{x}=-af(ax+by)+c_{1}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.11)

for some f​(t)βˆˆπŠβ€‹[t]π‘“π‘‘πŠdelimited-[]𝑑f(t)\in{\bf K}[t] and f​(0)=0𝑓00f(0)=0, a,bβˆˆπŠβˆ—π‘Žπ‘superscript𝐊a,b\in{\bf K}^{*}, c1,c2∈𝐊subscript𝑐1subscript𝑐2𝐊c_{1},c_{2}\in{\bf K}. Then equation (4.7) has the following form:

(u1​x​z+u0​x)​(v1​y​z+v0​y)βˆ’(v1​x​z+v0​x)​(u1​y​z+u0​y)βˆ’hx​(2​u2​z+u1)βˆ’hy​v1=0​(4.12)subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣1𝑦𝑧subscript𝑣0𝑦subscript𝑣1π‘₯𝑧subscript𝑣0π‘₯subscript𝑒1𝑦𝑧subscript𝑒0𝑦subscriptβ„Žπ‘₯2subscript𝑒2𝑧subscript𝑒1subscriptβ„Žπ‘¦subscript𝑣104.12(u_{1x}z+u_{0x})(v_{1y}z+v_{0y})-(v_{1x}z+v_{0x})(u_{1y}z+u_{0y})-h_{x}(2u_{2}z+u_{1})-h_{y}v_{1}=0~{}~{}~{}(4.12)

Thus, we have the following equations

u1​x​v0​y+u0​x​v1​yβˆ’v1​x​u0​yβˆ’v0​x​u1​yβˆ’2​u2​hx=0(4.13)subscript𝑒1π‘₯subscript𝑣0𝑦subscript𝑒0π‘₯subscript𝑣1𝑦subscript𝑣1π‘₯subscript𝑒0𝑦subscript𝑣0π‘₯subscript𝑒1𝑦2subscript𝑒2subscriptβ„Žπ‘₯04.13u_{1x}v_{0y}+u_{0x}v_{1y}-v_{1x}u_{0y}-v_{0x}u_{1y}-2u_{2}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.13)

and

u0​x​v0​yβˆ’v0​x​u0​yβˆ’hx​u1βˆ’hy​v1=0(4.14)subscript𝑒0π‘₯subscript𝑣0𝑦subscript𝑣0π‘₯subscript𝑒0𝑦subscriptβ„Žπ‘₯subscript𝑒1subscriptβ„Žπ‘¦subscript𝑣104.14u_{0x}v_{0y}-v_{0x}u_{0y}-h_{x}u_{1}-h_{y}v_{1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.14)

by comparing the coefficients of z𝑧z and z0superscript𝑧0z^{0} of equation (4.12). It follows from equations (4.3) and (4.5) that

[(v1​x​z+v0​x)​hyβˆ’hx​(v1​y​z+v0​y)]​(2​u2​z+u1)+[hx​(u1​y​z+u0​y)βˆ’hy​(u1​x​z+u0​x)]​v1=0​(4.15)delimited-[]subscript𝑣1π‘₯𝑧subscript𝑣0π‘₯subscriptβ„Žπ‘¦subscriptβ„Žπ‘₯subscript𝑣1𝑦𝑧subscript𝑣0𝑦2subscript𝑒2𝑧subscript𝑒1delimited-[]subscriptβ„Žπ‘₯subscript𝑒1𝑦𝑧subscript𝑒0𝑦subscriptβ„Žπ‘¦subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣104.15[(v_{1x}z+v_{0x})h_{y}-h_{x}(v_{1y}z+v_{0y})](2u_{2}z+u_{1})+[h_{x}(u_{1y}z+u_{0y})-h_{y}(u_{1x}z+u_{0x})]v_{1}=0~{}~{}~{}(4.15)

Comparing the coefficients of z2,z,z0superscript𝑧2𝑧superscript𝑧0z^{2},z,z^{0} of equation (4.15)4.15(4.15), we have the following equations:

v1​x​hyβˆ’hx​v1​y=0(4.16)subscript𝑣1π‘₯subscriptβ„Žπ‘¦subscriptβ„Žπ‘₯subscript𝑣1𝑦04.16v_{1x}h_{y}-h_{x}v_{1y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.16)

and

(v0​x​hyβˆ’hx​v0​y)​2​u2+(v1​x​hyβˆ’hx​v1​y)​u1+(hx​u1​yβˆ’hy​u1​x)​v1=0(4.17)subscript𝑣0π‘₯subscriptβ„Žπ‘¦subscriptβ„Žπ‘₯subscript𝑣0𝑦2subscript𝑒2subscript𝑣1π‘₯subscriptβ„Žπ‘¦subscriptβ„Žπ‘₯subscript𝑣1𝑦subscript𝑒1subscriptβ„Žπ‘₯subscript𝑒1𝑦subscriptβ„Žπ‘¦subscript𝑒1π‘₯subscript𝑣104.17(v_{0x}h_{y}-h_{x}v_{0y})2u_{2}+(v_{1x}h_{y}-h_{x}v_{1y})u_{1}+(h_{x}u_{1y}-h_{y}u_{1x})v_{1}=0~{}~{}~{}~{}~{}~{}(4.17)

and

(v0​x​hyβˆ’hx​v0​y)​u1+(hx​u0​yβˆ’hy​u0​x)​v1=0(4.18)subscript𝑣0π‘₯subscriptβ„Žπ‘¦subscriptβ„Žπ‘₯subscript𝑣0𝑦subscript𝑒1subscriptβ„Žπ‘₯subscript𝑒0𝑦subscriptβ„Žπ‘¦subscript𝑒0π‘₯subscript𝑣104.18(v_{0x}h_{y}-h_{x}v_{0y})u_{1}+(h_{x}u_{0y}-h_{y}u_{0x})v_{1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.18)

It follows from equations (4.11) and (4.16) that a​fβ€²β‹…(b​hxβˆ’a​hy)=0β‹…π‘Žsuperscript𝑓′𝑏subscriptβ„Žπ‘₯π‘Žsubscriptβ„Žπ‘¦0af^{\prime}\cdot(bh_{x}-ah_{y})=0. Thus, we have a=0π‘Ž0a=0 or fβ€²=0superscript𝑓′0f^{\prime}=0 or b​hx=a​hy𝑏subscriptβ„Žπ‘₯π‘Žsubscriptβ„Žπ‘¦bh_{x}=ah_{y}.

(i) If a=0π‘Ž0a=0, then

u1=b​f​(b​y)+c2,v1=c1βˆˆπŠβˆ—(4.19)formulae-sequenceformulae-sequencesubscript𝑒1𝑏𝑓𝑏𝑦subscript𝑐2subscript𝑣1subscript𝑐1superscript𝐊4.19u_{1}=bf(by)+c_{2},~{}~{}~{}v_{1}=c_{1}\in{\bf K}^{*}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.19)

It follows from equation (4.13) that 2​u2​hx=βˆ’v0​x​u1​y2subscript𝑒2subscriptβ„Žπ‘₯subscript𝑣0π‘₯subscript𝑒1𝑦2u_{2}h_{x}=-v_{0x}u_{1y}. That is,

2​u2​hx=βˆ’b2​f′​(b​y)​v0​x.2subscript𝑒2subscriptβ„Žπ‘₯superscript𝑏2superscript𝑓′𝑏𝑦subscript𝑣0π‘₯2u_{2}h_{x}=-b^{2}f^{\prime}(by)v_{0x}.

Integrating with respect to xπ‘₯x of two sides of the above equation, we have

h=βˆ’b22​u2​f′​(b​y)​v0+c​(y)2​u2(4.20)β„Žsuperscript𝑏22subscript𝑒2superscript𝑓′𝑏𝑦subscript𝑣0𝑐𝑦2subscript𝑒24.20h=-\frac{b^{2}}{2u_{2}}f^{\prime}(by)v_{0}+\frac{c(y)}{2u_{2}}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.20)

for some c​(y)βˆˆπŠβ€‹[y]π‘π‘¦πŠdelimited-[]𝑦c(y)\in{\bf K}[y]. Substituting equations (4.19) and (4.20) to equation (4.17), we have the following equation:

v0​x​[βˆ’b3​f′′​(b​y)​v0+c′​(y)βˆ’b42​u2​v1β‹…(f′​(b​y))2]=0.subscript𝑣0π‘₯delimited-[]superscript𝑏3superscript𝑓′′𝑏𝑦subscript𝑣0superscript𝑐′𝑦⋅superscript𝑏42subscript𝑒2subscript𝑣1superscriptsuperscript𝑓′𝑏𝑦20v_{0x}[-b^{3}f^{\prime\prime}(by)v_{0}+c^{\prime}(y)-\frac{b^{4}}{2u_{2}}v_{1}\cdot(f^{\prime}(by))^{2}]=0.

Thus, we have v0​x=0subscript𝑣0π‘₯0v_{0x}=0 or f′′​(b​y)=0superscript𝑓′′𝑏𝑦0f^{\prime\prime}(by)=0 and c′​(y)=b42​u2​v1β‹…(f′​(b​y))2superscript𝑐′𝑦⋅superscript𝑏42subscript𝑒2subscript𝑣1superscriptsuperscript𝑓′𝑏𝑦2c^{\prime}(y)=\frac{b^{4}}{2u_{2}}v_{1}\cdot(f^{\prime}(by))^{2}.

If v0​x=0subscript𝑣0π‘₯0v_{0x}=0, then it follows from equations (4.13) and (4.19) that hx=0subscriptβ„Žπ‘₯0h_{x}=0. It follows from equation (4.18) that v1​hy​u0​x=0subscript𝑣1subscriptβ„Žπ‘¦subscript𝑒0π‘₯0v_{1}h_{y}u_{0x}=0. Thus, we have hy=0subscriptβ„Žπ‘¦0h_{y}=0 or u0​x=0subscript𝑒0π‘₯0u_{0x}=0. If u0​x=0subscript𝑒0π‘₯0u_{0x}=0, then it follows from equation (4.14) that hy=0subscriptβ„Žπ‘¦0h_{y}=0. If hy=0subscriptβ„Žπ‘¦0h_{y}=0, then h=0β„Ž0h=0 because h​(0,0)=0β„Ž000h(0,0)=0. Thus, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.

If f′′​(b​y)=0superscript𝑓′′𝑏𝑦0f^{\prime\prime}(by)=0, then f′​(b​y)∈𝐊superscriptπ‘“β€²π‘π‘¦πŠf^{\prime}(by)\in{\bf K} and c′​(y)=b42​u2​v1​(f′​(b​y))2∈𝐊superscript𝑐′𝑦superscript𝑏42subscript𝑒2subscript𝑣1superscriptsuperscript𝑓′𝑏𝑦2𝐊c^{\prime}(y)=\frac{b^{4}}{2u_{2}}v_{1}(f^{\prime}(by))^{2}\in{\bf K}. Let l:=b2​f′​(b​y)assign𝑙superscript𝑏2superscript𝑓′𝑏𝑦l:=b^{2}f^{\prime}(by) and c:=l2(2​u2)2​v1assign𝑐superscript𝑙2superscript2subscript𝑒22subscript𝑣1c:=\frac{l^{2}}{(2u_{2})^{2}}v_{1}. Then it follows from equation (4.19) that v1βˆˆπŠβˆ—subscript𝑣1superscript𝐊v_{1}\in{\bf K}^{*} and u1=l​y+c2subscript𝑒1𝑙𝑦subscript𝑐2u_{1}=ly+c_{2} for some c2∈𝐊subscript𝑐2𝐊c_{2}\in{\bf K}. Since h​(0,0)=0β„Ž000h(0,0)=0, so it follows from equation (4.20) that

h=βˆ’l2​u2​v0+cβ‹…y(4.21)β„Žπ‘™2subscript𝑒2subscript𝑣0⋅𝑐𝑦4.21h=-\frac{l}{2u_{2}}v_{0}+c\cdot y~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.21)

It follows from equations (4.14) and (4.21) that

v0​x​u0​yβˆ’u0​x​v0​y=l2​u2​u1​v0​x+l2​u2​v1​v0​yβˆ’c​v1(4.22)subscript𝑣0π‘₯subscript𝑒0𝑦subscript𝑒0π‘₯subscript𝑣0𝑦𝑙2subscript𝑒2subscript𝑒1subscript𝑣0π‘₯𝑙2subscript𝑒2subscript𝑣1subscript𝑣0𝑦𝑐subscript𝑣14.22v_{0x}u_{0y}-u_{0x}v_{0y}=\frac{l}{2u_{2}}u_{1}v_{0x}+\frac{l}{2u_{2}}v_{1}v_{0y}-cv_{1}~{}~{}~{}~{}~{}~{}~{}~{}(4.22)

It follows from equations (4.18) and (4.21) that

cβ‹…u1​v0​x+v1​[βˆ’l2​u2​(v0​x​u0​yβˆ’u0​x​v0​y)βˆ’cβ‹…u0​x]=0(4.23)⋅𝑐subscript𝑒1subscript𝑣0π‘₯subscript𝑣1delimited-[]𝑙2subscript𝑒2subscript𝑣0π‘₯subscript𝑒0𝑦subscript𝑒0π‘₯subscript𝑣0𝑦⋅𝑐subscript𝑒0π‘₯04.23c\cdot u_{1}v_{0x}+v_{1}[-\frac{l}{2u_{2}}(v_{0x}u_{0y}-u_{0x}v_{0y})-c\cdot u_{0x}]=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.23)

Substituting equation (4.22) to equation (4.23), we have the following equation:

cβ‹…u1​v0​x+v1​[βˆ’l2(2​u2)2​u1​v0​x+l​c2​u2​v1βˆ’l2(2​u2)2​v1​v0​yβˆ’cβ‹…u0​x]=0.⋅𝑐subscript𝑒1subscript𝑣0π‘₯subscript𝑣1delimited-[]superscript𝑙2superscript2subscript𝑒22subscript𝑒1subscript𝑣0π‘₯𝑙𝑐2subscript𝑒2subscript𝑣1superscript𝑙2superscript2subscript𝑒22subscript𝑣1subscript𝑣0𝑦⋅𝑐subscript𝑒0π‘₯0c\cdot u_{1}v_{0x}+v_{1}[-\frac{l^{2}}{(2u_{2})^{2}}u_{1}v_{0x}+\frac{lc}{2u_{2}}v_{1}-\frac{l^{2}}{(2u_{2})^{2}}v_{1}v_{0y}-c\cdot u_{0x}]=0.

Since c=l2(2​u2)2​v1𝑐superscript𝑙2superscript2subscript𝑒22subscript𝑣1c=\frac{l^{2}}{(2u_{2})^{2}}v_{1}, so the above equation has the following form:

l​c2​u2​v1βˆ’c​(v0​y+u0​x)=0.𝑙𝑐2subscript𝑒2subscript𝑣1𝑐subscript𝑣0𝑦subscript𝑒0π‘₯0\frac{lc}{2u_{2}}v_{1}-c(v_{0y}+u_{0x})=0.

Substituting equation (4.6)(i=0)𝑖0(i=0) to the above equation, we have l​c2​u2​v1=0𝑙𝑐2subscript𝑒2subscript𝑣10\frac{lc}{2u_{2}}v_{1}=0. That is, l​c=0𝑙𝑐0lc=0. Since c=l2(2​u2)2​v1𝑐superscript𝑙2superscript2subscript𝑒22subscript𝑣1c=\frac{l^{2}}{(2u_{2})^{2}}v_{1}, so we have c=l=0𝑐𝑙0c=l=0. It follows from equation (4.21) that h=0β„Ž0h=0. Thus, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.

(ii) If fβ€²=0superscript𝑓′0f^{\prime}=0, then f=0𝑓0f=0 because f​(0)=0𝑓00f(0)=0. That is, u1=c2subscript𝑒1subscript𝑐2u_{1}=c_{2}, v1=c1βˆˆπŠβˆ—subscript𝑣1subscript𝑐1superscript𝐊v_{1}=c_{1}\in{\bf K}^{*}. It follows from equation (4.13) that hx=0subscriptβ„Žπ‘₯0h_{x}=0. It follows from equation (4.17) that v0​x​hy=0subscript𝑣0π‘₯subscriptβ„Žπ‘¦0v_{0x}h_{y}=0. Thus, we have hy=0subscriptβ„Žπ‘¦0h_{y}=0 or v0​x=0subscript𝑣0π‘₯0v_{0x}=0.
If hy=0subscriptβ„Žπ‘¦0h_{y}=0, then h=0β„Ž0h=0 because h​(0,0)=0β„Ž000h(0,0)=0. Thus, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.
If v0​x=0subscript𝑣0π‘₯0v_{0x}=0, then it follows from equation (4.18) that u0​x​hy=0subscript𝑒0π‘₯subscriptβ„Žπ‘¦0u_{0x}h_{y}=0. That is, u0​x=0subscript𝑒0π‘₯0u_{0x}=0 or hy=0subscriptβ„Žπ‘¦0h_{y}=0. If u0​x=0subscript𝑒0π‘₯0u_{0x}=0, then it follows from equation (4.14) that hy=0subscriptβ„Žπ‘¦0h_{y}=0. Thus, we have that hy=0subscriptβ„Žπ‘¦0h_{y}=0. It reduces to the above case.

(iii) If b​hx=a​hy𝑏subscriptβ„Žπ‘₯π‘Žsubscriptβ„Žπ‘¦bh_{x}=ah_{y}, then we have

hy=ba​hx(4.24)subscriptβ„Žπ‘¦π‘π‘Žsubscriptβ„Žπ‘₯4.24h_{y}=\frac{b}{a}h_{x}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.24)

because we can assume that aβ‹…fβ€²β‰ 0β‹…π‘Žsuperscript𝑓′0a\cdot f^{\prime}\neq 0. Otherwise, it reduces to the former two cases. Let xΒ―=a​x+b​yΒ―π‘₯π‘Žπ‘₯𝑏𝑦\bar{x}=ax+by, yΒ―=y¯𝑦𝑦\bar{y}=y. Then we have that hyΒ―=0subscriptβ„ŽΒ―π‘¦0h_{\bar{y}}=0. That is, hβˆˆπŠβ€‹[a​x+b​y]β„ŽπŠdelimited-[]π‘Žπ‘₯𝑏𝑦h\in{\bf K}[ax+by]. It follows from equations (4.10), (4.11), (4.17),(4.24) that

hx​(ba​v0​xβˆ’v0​y)=0(4.25)subscriptβ„Žπ‘₯π‘π‘Žsubscript𝑣0π‘₯subscript𝑣0𝑦04.25h_{x}(\frac{b}{a}v_{0x}-v_{0y})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.25)

It follows from equations (4.18) and (4.24) that

u1​hx​(ba​v0​xβˆ’v0​y)+v1​hx​(u0​yβˆ’ba​u0​x)=0(4.26)subscript𝑒1subscriptβ„Žπ‘₯π‘π‘Žsubscript𝑣0π‘₯subscript𝑣0𝑦subscript𝑣1subscriptβ„Žπ‘₯subscript𝑒0π‘¦π‘π‘Žsubscript𝑒0π‘₯04.26u_{1}h_{x}(\frac{b}{a}v_{0x}-v_{0y})+v_{1}h_{x}(u_{0y}-\frac{b}{a}u_{0x})=0~{}~{}~{}~{}~{}~{}~{}(4.26)

It follows from equations (4.25) and (4.26) that hx=0subscriptβ„Žπ‘₯0h_{x}=0 or b​v0​x=a​v0​y𝑏subscript𝑣0π‘₯π‘Žsubscript𝑣0𝑦bv_{0x}=av_{0y} and b​u0​x=a​u0​y𝑏subscript𝑒0π‘₯π‘Žsubscript𝑒0𝑦bu_{0x}=au_{0y}.
If hx=0subscriptβ„Žπ‘₯0h_{x}=0, then it follows from equation (4.24) that hy=0subscriptβ„Žπ‘¦0h_{y}=0. Thus, we have h=0β„Ž0h=0 because h​(0,0)=0β„Ž000h(0,0)=0. so u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.
If b​v0​x=a​v0​y𝑏subscript𝑣0π‘₯π‘Žsubscript𝑣0𝑦bv_{0x}=av_{0y} and b​u0​x=a​u0​y𝑏subscript𝑒0π‘₯π‘Žsubscript𝑒0𝑦bu_{0x}=au_{0y}, then v0,u0βˆˆπŠβ€‹[a​x+b​y]subscript𝑣0subscript𝑒0𝐊delimited-[]π‘Žπ‘₯𝑏𝑦v_{0},u_{0}\in{\bf K}[ax+by]. Thus, it follows from equations (4.10), (4.11) and (4.13) that hx=0subscriptβ„Žπ‘₯0h_{x}=0. It reduces to the former case.

(2) If v1​x=0subscript𝑣1π‘₯0v_{1x}=0, then it follows from equation (4.8) that

u1​x​v1​yβˆ’v0​x​u2​y=0(4.27)subscript𝑒1π‘₯subscript𝑣1𝑦subscript𝑣0π‘₯subscript𝑒2𝑦04.27u_{1x}v_{1y}-v_{0x}u_{2y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.27)

It follows from equation (4.7) that

(u1​x​z+u0​x)​(v1​y​z+v0​y)βˆ’v0​x​(u2​y​z2+u1​y​z+u0​y)βˆ’hx​(2​u2​z+u1)βˆ’hy​v1=0(4.28)subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣1𝑦𝑧subscript𝑣0𝑦subscript𝑣0π‘₯subscript𝑒2𝑦superscript𝑧2subscript𝑒1𝑦𝑧subscript𝑒0𝑦subscriptβ„Žπ‘₯2subscript𝑒2𝑧subscript𝑒1subscriptβ„Žπ‘¦subscript𝑣104.28(u_{1x}z+u_{0x})(v_{1y}z+v_{0y})-v_{0x}(u_{2y}z^{2}+u_{1y}z+u_{0y})-h_{x}(2u_{2}z+u_{1})-h_{y}v_{1}=0~{}~{}~{}~{}(4.28)

Comparing the coefficients of z2,z,z0superscript𝑧2𝑧superscript𝑧0z^{2},~{}z,~{}z^{0} of equation (4.28), we have the following equations:

{u1​x​v1​yβˆ’v0​x​u2​y=0(4.29)u1​x​v0​y+u0​x​v1​yβˆ’v0​x​u1​yβˆ’2​u2​hx=0(4.30)u0​x​v0​yβˆ’v0​x​u0​yβˆ’u1​hxβˆ’v1​hy=0(4.31)\left\{\begin{aligned} u_{1x}v_{1y}-v_{0x}u_{2y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.29)\\ u_{1x}v_{0y}+u_{0x}v_{1y}-v_{0x}u_{1y}-2u_{2}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}(4.30)\\ u_{0x}v_{0y}-v_{0x}u_{0y}-u_{1}h_{x}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.31)\\ \end{aligned}\right.

It follows from equations (4.3) and (4.5) that

[v0​x​hyβˆ’(v1​y​z+v0​y)​hx]​(2​u2​z+u1)+[hx​(u2​y​z2+u1​y​z+u0​y)βˆ’hy​(u1​x​z+u0​x)]​v1=0delimited-[]subscript𝑣0π‘₯subscriptβ„Žπ‘¦subscript𝑣1𝑦𝑧subscript𝑣0𝑦subscriptβ„Žπ‘₯2subscript𝑒2𝑧subscript𝑒1delimited-[]subscriptβ„Žπ‘₯subscript𝑒2𝑦superscript𝑧2subscript𝑒1𝑦𝑧subscript𝑒0𝑦subscriptβ„Žπ‘¦subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣10[v_{0x}h_{y}-(v_{1y}z+v_{0y})h_{x}](2u_{2}z+u_{1})+[h_{x}(u_{2y}z^{2}+u_{1y}z+u_{0y})-h_{y}(u_{1x}z+u_{0x})]v_{1}=0

Comparing the coefficients of z2,zsuperscript𝑧2𝑧z^{2},~{}z and z0superscript𝑧0z^{0} of the above equation, we have the following equations:

{hx​(v1​u2​yβˆ’2​u2​v1​y)=0(4.32)βˆ’v1​y​hx​u1+2​u2​(v0​x​hyβˆ’v0​y​hx)+v1​(hx​u1​yβˆ’hy​u1​x)=0(4.33)u1​(v0​x​hyβˆ’v0​y​hx)+v1​(hx​u0​yβˆ’hy​u0​x)=0(4.34)\left\{\begin{aligned} h_{x}(v_{1}u_{2y}-2u_{2}v_{1y})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.32)\\ -v_{1y}h_{x}u_{1}+2u_{2}(v_{0x}h_{y}-v_{0y}h_{x})+v_{1}(h_{x}u_{1y}-h_{y}u_{1x})=0~{}~{}~{}~{}(4.33)\\ u_{1}(v_{0x}h_{y}-v_{0y}h_{x})+v_{1}(h_{x}u_{0y}-h_{y}u_{0x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.34)\\ \end{aligned}\right.

It follows from equation (4.32) that hx=0subscriptβ„Žπ‘₯0h_{x}=0 or v1​u2​y=2​u2​v1​ysubscript𝑣1subscript𝑒2𝑦2subscript𝑒2subscript𝑣1𝑦v_{1}u_{2y}=2u_{2}v_{1y}.

If hx=0subscriptβ„Žπ‘₯0h_{x}=0, then it follows from equation (4.33) that hy​(2​u2​v0​xβˆ’u1​x​v1)=0subscriptβ„Žπ‘¦2subscript𝑒2subscript𝑣0π‘₯subscript𝑒1π‘₯subscript𝑣10h_{y}(2u_{2}v_{0x}-u_{1x}v_{1})=0. Thus, we have that hy=0subscriptβ„Žπ‘¦0h_{y}=0 or 2​u2​v0​x=v1​u1​x2subscript𝑒2subscript𝑣0π‘₯subscript𝑣1subscript𝑒1π‘₯2u_{2}v_{0x}=v_{1}u_{1x}. If hy=0subscriptβ„Žπ‘¦0h_{y}=0, then h=0β„Ž0h=0 because h​(0,0)=0β„Ž000h(0,0)=0. Thus, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent. If 2​u2​v0​x=v1​u1​x2subscript𝑒2subscript𝑣0π‘₯subscript𝑣1subscript𝑒1π‘₯2u_{2}v_{0x}=v_{1}u_{1x}, then it follows from equation (4.6) that

2​u2​v0​x=βˆ’v1​v1​yβˆˆπŠβ€‹[y](4.35)formulae-sequence2subscript𝑒2subscript𝑣0π‘₯subscript𝑣1subscript𝑣1π‘¦πŠdelimited-[]𝑦4.352u_{2}v_{0x}=-v_{1}v_{1y}\in{\bf K}[y]~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.35)

Substituting equations (4.35) and (4.6) to equation (4.29), we have the following equation: v1​y​(2​u2​v1​yβˆ’v1​u2​y)=0subscript𝑣1𝑦2subscript𝑒2subscript𝑣1𝑦subscript𝑣1subscript𝑒2𝑦0v_{1y}(2u_{2}v_{1y}-v_{1}u_{2y})=0. Thus, we have v1​y=0subscript𝑣1𝑦0v_{1y}=0 or 2​u2​v1​y=v1​u2​y2subscript𝑒2subscript𝑣1𝑦subscript𝑣1subscript𝑒2𝑦2u_{2}v_{1y}=v_{1}u_{2y}.
If v1​y=0subscript𝑣1𝑦0v_{1y}=0, then it follows from equation (4.29) that v0​x​u2​y=0subscript𝑣0π‘₯subscript𝑒2𝑦0v_{0x}u_{2y}=0. Thus, we have v0​x=0subscript𝑣0π‘₯0v_{0x}=0 or u2​y=0subscript𝑒2𝑦0u_{2y}=0. If u2​y=0subscript𝑒2𝑦0u_{2y}=0, then it reduces to Case I(1). If v0​x=0subscript𝑣0π‘₯0v_{0x}=0, then it follows form equation (4.34) that hy​u0​x=0subscriptβ„Žπ‘¦subscript𝑒0π‘₯0h_{y}u_{0x}=0. Thus, we have hy=0subscriptβ„Žπ‘¦0h_{y}=0 or u0​x=0subscript𝑒0π‘₯0u_{0x}=0. If u0​x=0subscript𝑒0π‘₯0u_{0x}=0, then it follows from equation (4.31) that hy=0subscriptβ„Žπ‘¦0h_{y}=0. Therefore, we have h=0β„Ž0h=0 because h​(0,0)=0β„Ž000h(0,0)=0. So u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.

If 2​u2​v1​y=v1​u2​y2subscript𝑒2subscript𝑣1𝑦subscript𝑣1subscript𝑒2𝑦2u_{2}v_{1y}=v_{1}u_{2y}, then we have

u2​yu2=2​v1​yv1(4.36)subscript𝑒2𝑦subscript𝑒22subscript𝑣1𝑦subscript𝑣14.36\frac{u_{2y}}{u_{2}}=2\frac{v_{1y}}{v_{1}}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.36)

Suppose u2​y​v1​yβ‰ 0subscript𝑒2𝑦subscript𝑣1𝑦0u_{2y}v_{1y}\neq 0. Then we have

u2=ec¯​(x)​v12subscript𝑒2superscript𝑒¯𝑐π‘₯superscriptsubscript𝑣12u_{2}=e^{\bar{c}(x)}v_{1}^{2}

by integrating of two sides of equation (4.36) with respect to y𝑦y. where c¯​(x)¯𝑐π‘₯\bar{c}(x) is a function of xπ‘₯x. Since u2,v1βˆˆπŠβ€‹[x,y]subscript𝑒2subscript𝑣1𝐊π‘₯𝑦u_{2},v_{1}\in{\bf K}[x,y], we have ec¯​(x)βˆˆπŠβ€‹[x]superscript𝑒¯𝑐π‘₯𝐊delimited-[]π‘₯e^{\bar{c}(x)}\in{\bf K}[x]. That is, u2=c​(x)​v12subscript𝑒2𝑐π‘₯superscriptsubscript𝑣12u_{2}=c(x)v_{1}^{2} for c​(x)βˆˆπŠβ€‹[x]𝑐π‘₯𝐊delimited-[]π‘₯c(x)\in{\bf K}[x] and c​(x)β‰ 0𝑐π‘₯0c(x)\neq 0. Then it follows from equations (4.29) and (4.6) that

v1​y​(2​c​(x)​v1​v0​x+v1​y)=0.subscript𝑣1𝑦2𝑐π‘₯subscript𝑣1subscript𝑣0π‘₯subscript𝑣1𝑦0v_{1y}(2c(x)v_{1}v_{0x}+v_{1y})=0.

That is,

2​c​(x)​v1​v0​x=βˆ’v1​y(4.37)2𝑐π‘₯subscript𝑣1subscript𝑣0π‘₯subscript𝑣1𝑦4.372c(x)v_{1}v_{0x}=-v_{1y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.37)

If v0​xβ‰ 0subscript𝑣0π‘₯0v_{0x}\neq 0, then we have that v1​y=0subscript𝑣1𝑦0v_{1y}=0 by comparing the degree of y𝑦y of equation (4.37). Thus, we have v0​x=0subscript𝑣0π‘₯0v_{0x}=0. This is a contradiction. Therefore, we have v1​y=v0​x=0subscript𝑣1𝑦subscript𝑣0π‘₯0v_{1y}=v_{0x}=0. It follows from equation (4.36) that u2​y=0subscript𝑒2𝑦0u_{2y}=0, This is a contradiction! If u2​y​v1​y=0subscript𝑒2𝑦subscript𝑣1𝑦0u_{2y}v_{1y}=0, then it follows from equation (4.36) that u2​y=v1​y=0subscript𝑒2𝑦subscript𝑣1𝑦0u_{2y}=v_{1y}=0. which reduces to case I(1).

Case II If dβ‰₯3𝑑3d\geq 3, then we have

ud​y​v1​x=0(4.38)subscript𝑒𝑑𝑦subscript𝑣1π‘₯04.38u_{dy}v_{1x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.38)

and

v1​x​u(dβˆ’1)​y+v0​x​ud​y=0(4.39)subscript𝑣1π‘₯subscript𝑒𝑑1𝑦subscript𝑣0π‘₯subscript𝑒𝑑𝑦04.39v_{1x}u_{(d-1)y}+v_{0x}u_{dy}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.39)

by comparing the coefficients of zd+1superscript𝑧𝑑1z^{d+1} and zdsuperscript𝑧𝑑z^{d} of equation (4.7). It follows from equations (4.38) and (4.39) that v1​x=v0​x=0subscript𝑣1π‘₯subscript𝑣0π‘₯0v_{1x}=v_{0x}=0 or v1​x=0=ud​ysubscript𝑣1π‘₯0subscript𝑒𝑑𝑦v_{1x}=0=u_{dy} or ud​y=u(dβˆ’1)​y=0subscript𝑒𝑑𝑦subscript𝑒𝑑1𝑦0u_{dy}=u_{(d-1)y}=0.

(a) If v1​x=v0​x=0subscript𝑣1π‘₯subscript𝑣0π‘₯0v_{1x}=v_{0x}=0, then equation (4.7) has the following form:

(u1​x​z+u0​x)​(v1​y​z+v0​y)βˆ’hx​(d​ud​zdβˆ’1+β‹―+u1)βˆ’hy​v1=0(4.40)subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣1𝑦𝑧subscript𝑣0𝑦subscriptβ„Žπ‘₯𝑑subscript𝑒𝑑superscript𝑧𝑑1β‹―subscript𝑒1subscriptβ„Žπ‘¦subscript𝑣104.40(u_{1x}z+u_{0x})(v_{1y}z+v_{0y})-h_{x}(du_{d}z^{d-1}+\cdots+u_{1})-h_{y}v_{1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.40)

If d>3𝑑3d>3, then hx=0subscriptβ„Žπ‘₯0h_{x}=0 by comparing the coefficient of zdβˆ’1superscript𝑧𝑑1z^{d-1} of equation (4.40). Thus, it follows from equation (4.3) that hy​(u1​x​z+u0​x)=0subscriptβ„Žπ‘¦subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯0h_{y}(u_{1x}z+u_{0x})=0. Therefore, we have hy=0subscriptβ„Žπ‘¦0h_{y}=0 or u1​x=u0​x=0subscript𝑒1π‘₯subscript𝑒0π‘₯0u_{1x}=u_{0x}=0. If u1​x=u0​x=0subscript𝑒1π‘₯subscript𝑒0π‘₯0u_{1x}=u_{0x}=0, then it follows from equation (4.40) that hy=0subscriptβ„Žπ‘¦0h_{y}=0. Thus, we have h=0β„Ž0h=0 because h​(0,0)=0β„Ž000h(0,0)=0. Therefore, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.
If d=3𝑑3d=3, then comparing the coefficients of z2,zsuperscript𝑧2𝑧z^{2},z and z0superscript𝑧0z^{0} of equation (4.40), we have the following equations:

{u1​x​v1​yβˆ’3​u3​hx=0(4.41)u1​x​v0​yβˆ’v1​y​u0​xβˆ’2​u2​hx=0(4.42)u0​x​v0​yβˆ’u1​hxβˆ’v1​hy=0(4.43)\left\{\begin{aligned} u_{1x}v_{1y}-3u_{3}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.41)\\ u_{1x}v_{0y}-v_{1y}u_{0x}-2u_{2}h_{x}=0~{}~{}~{}~{}~{}~{}~{}(4.42)\\ u_{0x}v_{0y}-u_{1}h_{x}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.43)\\ \end{aligned}\right.

It follows from equations (4.3) and (4.5) that βˆ’hx​(v1​y​z+v0​y)​(3​u3​z2+2​u2​z+u1)+[hx​(u3​y​z3+u2​y​z2+u1​y​z+u0​y)βˆ’hy​(u1​x​z+u0​x)]​v1=0subscriptβ„Žπ‘₯subscript𝑣1𝑦𝑧subscript𝑣0𝑦3subscript𝑒3superscript𝑧22subscript𝑒2𝑧subscript𝑒1delimited-[]subscriptβ„Žπ‘₯subscript𝑒3𝑦superscript𝑧3subscript𝑒2𝑦superscript𝑧2subscript𝑒1𝑦𝑧subscript𝑒0𝑦subscriptβ„Žπ‘¦subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣10-h_{x}(v_{1y}z+v_{0y})(3u_{3}z^{2}+2u_{2}z+u_{1})+[h_{x}(u_{3y}z^{3}+u_{2y}z^{2}+u_{1y}z+u_{0y})-h_{y}(u_{1x}z+u_{0x})]v_{1}=0 Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (4.44)
Comparing the coefficients of z3superscript𝑧3z^{3} of the above equation, we have hx​(3​v1​y​u3βˆ’u3​y​v1)=0subscriptβ„Žπ‘₯3subscript𝑣1𝑦subscript𝑒3subscript𝑒3𝑦subscript𝑣10h_{x}(3v_{1y}u_{3}-u_{3y}v_{1})=0. Thus, we have hx=0subscriptβ„Žπ‘₯0h_{x}=0 or 3​u3​v1​y=v1​u3​y3subscript𝑒3subscript𝑣1𝑦subscript𝑣1subscript𝑒3𝑦3u_{3}v_{1y}=v_{1}u_{3y}.

(1) If hx=0subscriptβ„Žπ‘₯0h_{x}=0, then it follows from equation (4.44) that hy​(u1​x​z+u0​x)=0subscriptβ„Žπ‘¦subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯0h_{y}(u_{1x}z+u_{0x})=0. That is, hy=0subscriptβ„Žπ‘¦0h_{y}=0 or u1​x=u0​x=0subscript𝑒1π‘₯subscript𝑒0π‘₯0u_{1x}=u_{0x}=0. If u1​x=u0​x=0subscript𝑒1π‘₯subscript𝑒0π‘₯0u_{1x}=u_{0x}=0, then it follows from equation (4.40) that hy=0subscriptβ„Žπ‘¦0h_{y}=0. Thus, we have h=0β„Ž0h=0 because h​(0,0)=0β„Ž000h(0,0)=0. Therefore, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.

(2) If 3​u3​v1​y=v1​u3​y3subscript𝑒3subscript𝑣1𝑦subscript𝑣1subscript𝑒3𝑦3u_{3}v_{1y}=v_{1}u_{3y}, then

u3​yu3=3​v1​yv1(4.45)subscript𝑒3𝑦subscript𝑒33subscript𝑣1𝑦subscript𝑣14.45\frac{u_{3y}}{u_{3}}=3\frac{v_{1y}}{v_{1}}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.45)

If v1​y=0subscript𝑣1𝑦0v_{1y}=0, then u3​y=0subscript𝑒3𝑦0u_{3y}=0. It follows from equation (4.41) that hx=0subscriptβ„Žπ‘₯0h_{x}=0. Then it follows from the arguments of Case II (1) that u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent. We can assume that u3​y​v1​yβ‰ 0subscript𝑒3𝑦subscript𝑣1𝑦0u_{3y}v_{1y}\neq 0. Then we have that u3=ed¯​(x)​v13subscript𝑒3superscript𝑒¯𝑑π‘₯superscriptsubscript𝑣13u_{3}=e^{\bar{d}(x)}v_{1}^{3} by integrating the two sides of equation (4.45) with respect to y𝑦y, where d¯​(x)¯𝑑π‘₯\bar{d}(x) is a function of xπ‘₯x. Since u3,v1βˆˆπŠβ€‹[x,y]subscript𝑒3subscript𝑣1𝐊π‘₯𝑦u_{3},v_{1}\in{\bf K}[x,y], we have ed¯​(x)βˆˆπŠβ€‹[x]superscript𝑒¯𝑑π‘₯𝐊delimited-[]π‘₯e^{\bar{d}(x)}\in{\bf K}[x]. That is,

u3=d​(x)​v13(4.46)subscript𝑒3𝑑π‘₯superscriptsubscript𝑣134.46u_{3}=d(x)v_{1}^{3}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.46)

with d​(x)βˆˆπŠβ€‹[x]𝑑π‘₯𝐊delimited-[]π‘₯d(x)\in{\bf K}[x], d​(x)β‰ 0𝑑π‘₯0d(x)\neq 0. Substituting equations (4.6) and (4.46)4.46(4.46) to equation (4.41), we have that

βˆ’3​d​(x)​v13​hx=v1​y2(4.47)3𝑑π‘₯superscriptsubscript𝑣13subscriptβ„Žπ‘₯superscriptsubscript𝑣1𝑦24.47-3d(x)v_{1}^{3}h_{x}=v_{1y}^{2}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.47)

Then we have v1​y=0subscript𝑣1𝑦0v_{1y}=0 by comparing the degree of y𝑦y of equation (4.47). It follows from equation (4.47) that hx=0subscriptβ„Žπ‘₯0h_{x}=0. Then it follows from the arguments of Case II(1) that u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.

(b) If v1​x=ud​y=0subscript𝑣1π‘₯subscript𝑒𝑑𝑦0v_{1x}=u_{dy}=0, then it follows from equation (4.6) that udβˆˆπŠβˆ—subscript𝑒𝑑superscript𝐊u_{d}\in{\bf K}^{*}. then equation (4.7) has the following form: (u1​x​z+u0​x)​(v1​y​z+v0​y)βˆ’v0​x​(u(dβˆ’1)​y​zdβˆ’1+β‹―+u1​y​z+u0​y)βˆ’hx​(d​ud​zdβˆ’1+β‹―+u1)βˆ’hy​v1=0subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣1𝑦𝑧subscript𝑣0𝑦subscript𝑣0π‘₯subscript𝑒𝑑1𝑦superscript𝑧𝑑1β‹―subscript𝑒1𝑦𝑧subscript𝑒0𝑦subscriptβ„Žπ‘₯𝑑subscript𝑒𝑑superscript𝑧𝑑1β‹―subscript𝑒1subscriptβ„Žπ‘¦subscript𝑣10(u_{1x}z+u_{0x})(v_{1y}z+v_{0y})-v_{0x}(u_{(d-1)y}z^{d-1}+\cdots+u_{1y}z+u_{0y})-h_{x}(du_{d}z^{d-1}+\cdots+u_{1})-h_{y}v_{1}=0 Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (4.48)
Then we have the following equations:

βˆ’v0​x​ui​yβˆ’(i+1)​ui+1​hx=0(4.49)subscript𝑣0π‘₯subscript𝑒𝑖𝑦𝑖1subscript𝑒𝑖1subscriptβ„Žπ‘₯04.49-v_{0x}u_{iy}-(i+1)u_{i+1}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.49)

by comparing the coefficients of zisuperscript𝑧𝑖z^{i} of equation (4.49) for i=dβˆ’1,dβˆ’2,…,3𝑖𝑑1𝑑2…3i=d-1,d-2,\ldots,3. Comparing the coefficients of z2,zsuperscript𝑧2𝑧z^{2},z and z0superscript𝑧0z^{0} of equation (4.48), we have the following equations:

{u1​x​v1​yβˆ’v0​x​u2​yβˆ’3​u3​hx=0(4.50)u1​x​v0​y+v1​y​u0​xβˆ’v0​x​u1​yβˆ’2​u2​hx=0(4.51)u0​x​v0​yβˆ’v0​x​u0​yβˆ’u1​hxβˆ’v1​hy=0(4.52)\left\{\begin{aligned} u_{1x}v_{1y}-v_{0x}u_{2y}-3u_{3}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.50)\\ u_{1x}v_{0y}+v_{1y}u_{0x}-v_{0x}u_{1y}-2u_{2}h_{x}=0~{}~{}~{}~{}~{}~{}~{}(4.51)\\ u_{0x}v_{0y}-v_{0x}u_{0y}-u_{1}h_{x}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.52)\\ \end{aligned}\right.

It follows from equations (4.3) and (4.5) that
[v0​x​hyβˆ’hx​(v1​y​z+v0​y)]​(d​ud​zdβˆ’1+(dβˆ’1)​udβˆ’1​zdβˆ’2+β‹―+u1)+[hx​(u(dβˆ’1)​y​zdβˆ’1+β‹―+u1​y​z+u0​y)βˆ’hy​(u1​x​z+u0​x)]​v1=0delimited-[]subscript𝑣0π‘₯subscriptβ„Žπ‘¦subscriptβ„Žπ‘₯subscript𝑣1𝑦𝑧subscript𝑣0𝑦𝑑subscript𝑒𝑑superscript𝑧𝑑1𝑑1subscript𝑒𝑑1superscript𝑧𝑑2β‹―subscript𝑒1delimited-[]subscriptβ„Žπ‘₯subscript𝑒𝑑1𝑦superscript𝑧𝑑1β‹―subscript𝑒1𝑦𝑧subscript𝑒0𝑦subscriptβ„Žπ‘¦subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣10[v_{0x}h_{y}-h_{x}(v_{1y}z+v_{0y})](du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})+[h_{x}(u_{(d-1)y}z^{d-1}+\cdots+u_{1y}z+u_{0y})-h_{y}(u_{1x}z+u_{0x})]v_{1}=0Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (4.53)
Then we have hx​v1​y=0subscriptβ„Žπ‘₯subscript𝑣1𝑦0h_{x}v_{1y}=0 by comparing the coefficients of zdsuperscript𝑧𝑑z^{d} of equation (4.53). That is, hx=0subscriptβ„Žπ‘₯0h_{x}=0 or v1​y=0subscript𝑣1𝑦0v_{1y}=0.

(3) If hx=0subscriptβ„Žπ‘₯0h_{x}=0, then equation (4.53) has the following form :

v0​x​hy​(d​ud​zdβˆ’1+(dβˆ’1)​udβˆ’1​zdβˆ’2+β‹―+u1)βˆ’hy​(u1​x​z+u0​x)​v1=0(4.54)subscript𝑣0π‘₯subscriptβ„Žπ‘¦π‘‘subscript𝑒𝑑superscript𝑧𝑑1𝑑1subscript𝑒𝑑1superscript𝑧𝑑2β‹―subscript𝑒1subscriptβ„Žπ‘¦subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣104.54v_{0x}h_{y}(du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})-h_{y}(u_{1x}z+u_{0x})v_{1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.54)

Comparing the coefficients of zdβˆ’1superscript𝑧𝑑1z^{d-1} of equation (4.54), we have that v0​x​hy=0subscript𝑣0π‘₯subscriptβ„Žπ‘¦0v_{0x}h_{y}=0. That is, v0​x=0subscript𝑣0π‘₯0v_{0x}=0 or hy=0subscriptβ„Žπ‘¦0h_{y}=0. If v0​x=0subscript𝑣0π‘₯0v_{0x}=0, then it reduces to Case II(a). If hy=0subscriptβ„Žπ‘¦0h_{y}=0, then h=0β„Ž0h=0 because h​(0,0)=0β„Ž000h(0,0)=0. Thus, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent.

(4) If v1​y=0subscript𝑣1𝑦0v_{1y}=0, then comparing the coefficients of zdβˆ’1superscript𝑧𝑑1z^{d-1} and z0superscript𝑧0z^{0} of equation (4.53), we have

(v0​x​hyβˆ’hx​v0​y)​d​ud+hx​u(dβˆ’1)​y​v1=0(4.55)subscript𝑣0π‘₯subscriptβ„Žπ‘¦subscriptβ„Žπ‘₯subscript𝑣0𝑦𝑑subscript𝑒𝑑subscriptβ„Žπ‘₯subscript𝑒𝑑1𝑦subscript𝑣104.55(v_{0x}h_{y}-h_{x}v_{0y})du_{d}+h_{x}u_{(d-1)y}v_{1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.55)

and

(v0​x​hyβˆ’hx​v0​y)​u1+(hx​u0​yβˆ’hy​u0​x)​v1=0(4.56)subscript𝑣0π‘₯subscriptβ„Žπ‘¦subscriptβ„Žπ‘₯subscript𝑣0𝑦subscript𝑒1subscriptβ„Žπ‘₯subscript𝑒0𝑦subscriptβ„Žπ‘¦subscript𝑒0π‘₯subscript𝑣104.56(v_{0x}h_{y}-h_{x}v_{0y})u_{1}+(h_{x}u_{0y}-h_{y}u_{0x})v_{1}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.56)

If d>3𝑑3d>3, then it follows from equation (4.49)(i=dβˆ’1)𝑖𝑑1(i=d-1) that

hx=βˆ’1d​ud​v0​x​u(dβˆ’1)​y(4.57)subscriptβ„Žπ‘₯1𝑑subscript𝑒𝑑subscript𝑣0π‘₯subscript𝑒𝑑1𝑦4.57h_{x}=-\frac{1}{du_{d}}v_{0x}u_{(d-1)y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.57)

If d=3𝑑3d=3, then it follows from equation (4.50) that

hx=βˆ’13​u3​v0​x​u2​y(4.57)subscriptβ„Žπ‘₯13subscript𝑒3subscript𝑣0π‘₯subscript𝑒2𝑦4.57h_{x}=-\frac{1}{3u_{3}}v_{0x}u_{2y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.57)

Substituting equation (4.57) to equation (4.55), we have

v0​x​[hyβˆ’v1d2​ud2​u(dβˆ’1)​y2+1d​ud​u(dβˆ’1)​y​v0​y]=0subscript𝑣0π‘₯delimited-[]subscriptβ„Žπ‘¦subscript𝑣1superscript𝑑2superscriptsubscript𝑒𝑑2superscriptsubscript𝑒𝑑1𝑦21𝑑subscript𝑒𝑑subscript𝑒𝑑1𝑦subscript𝑣0𝑦0v_{0x}[h_{y}-\frac{v_{1}}{d^{2}u_{d}^{2}}u_{(d-1)y}^{2}+\frac{1}{du_{d}}u_{(d-1)y}v_{0y}]=0

for dβ‰₯3𝑑3d\geq 3. Thus, we have v0​x=0subscript𝑣0π‘₯0v_{0x}=0 or

hy=v1d2​ud2​u(dβˆ’1)​y2βˆ’1d​ud​u(dβˆ’1)​y​v0​y(4.58)subscriptβ„Žπ‘¦subscript𝑣1superscript𝑑2superscriptsubscript𝑒𝑑2superscriptsubscript𝑒𝑑1𝑦21𝑑subscript𝑒𝑑subscript𝑒𝑑1𝑦subscript𝑣0𝑦4.58h_{y}=\frac{v_{1}}{d^{2}u_{d}^{2}}u_{(d-1)y}^{2}-\frac{1}{du_{d}}u_{(d-1)y}v_{0y}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.58)

If v0​x=0subscript𝑣0π‘₯0v_{0x}=0, then it reduces to Case II(a). Otherwise, substituting equations (4.57) and (4.58) to equation (4.52),we have that

u0​x​v0​yβˆ’v0​x​u0​y=βˆ’u1d​ud​v0​x​u(dβˆ’1)​y+v12d2​ud2​u(dβˆ’1)​y2βˆ’v1d​ud​u(dβˆ’1)​y​v0​y(4.59)subscript𝑒0π‘₯subscript𝑣0𝑦subscript𝑣0π‘₯subscript𝑒0𝑦subscript𝑒1𝑑subscript𝑒𝑑subscript𝑣0π‘₯subscript𝑒𝑑1𝑦superscriptsubscript𝑣12superscript𝑑2superscriptsubscript𝑒𝑑2superscriptsubscript𝑒𝑑1𝑦2subscript𝑣1𝑑subscript𝑒𝑑subscript𝑒𝑑1𝑦subscript𝑣0𝑦4.59u_{0x}v_{0y}-v_{0x}u_{0y}=-\frac{u_{1}}{du_{d}}v_{0x}u_{(d-1)y}+\frac{v_{1}^{2}}{d^{2}u_{d}^{2}}u_{(d-1)y}^{2}-\frac{v_{1}}{du_{d}}u_{(d-1)y}v_{0y}~{}~{}~{}~{}~{}~{}~{}(4.59)

Substituting equations (4.57) and (4.58) to equation (4.56), we have that

u1​v1d2​ud2​u(dβˆ’1)​y2​v0​xβˆ’v12d2​ud2​u(dβˆ’1)​y2​u0​x+v1d​ud​u(dβˆ’1)​y​(u0​x​v0​yβˆ’v0​x​u0​y)=0(4.60)subscript𝑒1subscript𝑣1superscript𝑑2superscriptsubscript𝑒𝑑2superscriptsubscript𝑒𝑑1𝑦2subscript𝑣0π‘₯superscriptsubscript𝑣12superscript𝑑2superscriptsubscript𝑒𝑑2superscriptsubscript𝑒𝑑1𝑦2subscript𝑒0π‘₯subscript𝑣1𝑑subscript𝑒𝑑subscript𝑒𝑑1𝑦subscript𝑒0π‘₯subscript𝑣0𝑦subscript𝑣0π‘₯subscript𝑒0𝑦04.60\frac{u_{1}v_{1}}{d^{2}u_{d}^{2}}u_{(d-1)y}^{2}v_{0x}-\frac{v_{1}^{2}}{d^{2}u_{d}^{2}}u_{(d-1)y}^{2}u_{0x}+\frac{v_{1}}{du_{d}}u_{(d-1)y}(u_{0x}v_{0y}-v_{0x}u_{0y})=0~{}~{}~{}~{}~{}(4.60)

Then we have v13d3​ud3​u(dβˆ’1)​y3=0superscriptsubscript𝑣13superscript𝑑3superscriptsubscript𝑒𝑑3superscriptsubscript𝑒𝑑1𝑦30\frac{v_{1}^{3}}{d^{3}u_{d}^{3}}u_{(d-1)y}^{3}=0 by substituting equations (4.6) and (4.59) to equation (4.60). That is, u(dβˆ’1)​y=0subscript𝑒𝑑1𝑦0u_{(d-1)y}=0. It follows from equation (4.49)(i=dβˆ’1)𝑖𝑑1(i=d-1) that hx=0subscriptβ„Žπ‘₯0h_{x}=0. Then it reduces to Case II (3).

(c) If ud​y=u(dβˆ’1)​y=0subscript𝑒𝑑𝑦subscript𝑒𝑑1𝑦0u_{dy}=u_{(d-1)y}=0, then it follows from equations (4.2) and (4.5) that (u1​x​z+u0​x)​(v1​y​z+v0​y)βˆ’(v1​x​z+v0​x)​(u(dβˆ’2)​y​zdβˆ’2+β‹―+u1​y​z+u0​y)βˆ’hx​(d​ud​zdβˆ’1+β‹―+u1)βˆ’hy​v1=0subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣1𝑦𝑧subscript𝑣0𝑦subscript𝑣1π‘₯𝑧subscript𝑣0π‘₯subscript𝑒𝑑2𝑦superscript𝑧𝑑2β‹―subscript𝑒1𝑦𝑧subscript𝑒0𝑦subscriptβ„Žπ‘₯𝑑subscript𝑒𝑑superscript𝑧𝑑1β‹―subscript𝑒1subscriptβ„Žπ‘¦subscript𝑣10(u_{1x}z+u_{0x})(v_{1y}z+v_{0y})-(v_{1x}z+v_{0x})(u_{(d-2)y}z^{d-2}+\cdots+u_{1y}z+u_{0y})-h_{x}(du_{d}z^{d-1}+\cdots+u_{1})-h_{y}v_{1}=0 Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (4.61)
Comparing the coefficients of zjsuperscript𝑧𝑗z^{j} of equation (4.61) for j=dβˆ’2,…,3𝑗𝑑2…3j=d-2,\ldots,3, we have the following equations:

βˆ’v1​x​u(jβˆ’1)​yβˆ’v0​x​uj​yβˆ’(j+1)​uj+1​hx=0(4.62)subscript𝑣1π‘₯subscript𝑒𝑗1𝑦subscript𝑣0π‘₯subscript𝑒𝑗𝑦𝑗1subscript𝑒𝑗1subscriptβ„Žπ‘₯04.62-v_{1x}u_{(j-1)y}-v_{0x}u_{jy}-(j+1)u_{j+1}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.62)

Comparing the coefficients of zdβˆ’1,z2,zsuperscript𝑧𝑑1superscript𝑧2𝑧z^{d-1},z^{2},z and z0superscript𝑧0z^{0}, we have the following equations:

{βˆ’v1​x​u(dβˆ’2)​yβˆ’d​ud​hx=0(4.63)u1​x​v1​yβˆ’v1​x​u1​yβˆ’v0​x​u2​yβˆ’3​u3​hx=0(4.64)u1​x​v0​y+v1​y​u0​xβˆ’v1​x​u0​yβˆ’v0​x​u1​yβˆ’2​u2​hx=0(4.65)u0​x​v0​yβˆ’v0​x​u0​yβˆ’u1​hxβˆ’v1​hy=0(4.66)\left\{\begin{aligned} -v_{1x}u_{(d-2)y}-du_{d}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.63)\\ u_{1x}v_{1y}-v_{1x}u_{1y}-v_{0x}u_{2y}-3u_{3}h_{x}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.64)\\ u_{1x}v_{0y}+v_{1y}u_{0x}-v_{1x}u_{0y}-v_{0x}u_{1y}-2u_{2}h_{x}=0~{}~{}~{}~{}~{}~{}~{}(4.65)\\ u_{0x}v_{0y}-v_{0x}u_{0y}-u_{1}h_{x}-v_{1}h_{y}=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.66)\\ \end{aligned}\right.

It follows from equations (4.3) and (4.5) that
[(v1​x​z+v0​x)​hyβˆ’hx​(v1​y​z+v0​y)]​(d​ud​zdβˆ’1+(dβˆ’1)​udβˆ’1​zdβˆ’2+β‹―+u1)+[hx​(u(dβˆ’2)​y​zdβˆ’2+β‹―+u1​y​z+u0​y)βˆ’hy​(u1​x​z+u0​x)]​v1=0delimited-[]subscript𝑣1π‘₯𝑧subscript𝑣0π‘₯subscriptβ„Žπ‘¦subscriptβ„Žπ‘₯subscript𝑣1𝑦𝑧subscript𝑣0𝑦𝑑subscript𝑒𝑑superscript𝑧𝑑1𝑑1subscript𝑒𝑑1superscript𝑧𝑑2β‹―subscript𝑒1delimited-[]subscriptβ„Žπ‘₯subscript𝑒𝑑2𝑦superscript𝑧𝑑2β‹―subscript𝑒1𝑦𝑧subscript𝑒0𝑦subscriptβ„Žπ‘¦subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯subscript𝑣10[(v_{1x}z+v_{0x})h_{y}-h_{x}(v_{1y}z+v_{0y})](du_{d}z^{d-1}+(d-1)u_{d-1}z^{d-2}+\cdots+u_{1})+[h_{x}(u_{(d-2)y}z^{d-2}+\cdots+u_{1y}z+u_{0y})-h_{y}(u_{1x}z+u_{0x})]v_{1}=0 Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β (4.67)
Comparing the coefficients of zdsuperscript𝑧𝑑z^{d} and zdβˆ’1superscript𝑧𝑑1z^{d-1} of equation (4.67), we have the following equations:

{d​ud​(v1​x​hyβˆ’hx​v1​y)=0(dβˆ’1)​udβˆ’1​(v1​x​hyβˆ’hx​v1​y)+d​ud​(v0​x​hyβˆ’hx​v0​y)=0\left\{\begin{aligned} du_{d}(v_{1x}h_{y}-h_{x}v_{1y})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\ (d-1)u_{d-1}(v_{1x}h_{y}-h_{x}v_{1y})+du_{d}(v_{0x}h_{y}-h_{x}v_{0y})=0\\ \end{aligned}\right.

That is, v1​x​hyβˆ’hx​v1​y=0subscript𝑣1π‘₯subscriptβ„Žπ‘¦subscriptβ„Žπ‘₯subscript𝑣1𝑦0v_{1x}h_{y}-h_{x}v_{1y}=0 and v0​x​hyβˆ’hx​v0​y=0subscript𝑣0π‘₯subscriptβ„Žπ‘¦subscriptβ„Žπ‘₯subscript𝑣0𝑦0v_{0x}h_{y}-h_{x}v_{0y}=0. Then equation (4.67) has the following form :

hx​(u(dβˆ’2)​y​zdβˆ’2+β‹―+u1​y​z+u0​y)βˆ’hy​(u1​x​z+u0​x)=0(4.68)subscriptβ„Žπ‘₯subscript𝑒𝑑2𝑦superscript𝑧𝑑2β‹―subscript𝑒1𝑦𝑧subscript𝑒0𝑦subscriptβ„Žπ‘¦subscript𝑒1π‘₯𝑧subscript𝑒0π‘₯04.68h_{x}(u_{(d-2)y}z^{d-2}+\cdots+u_{1y}z+u_{0y})-h_{y}(u_{1x}z+u_{0x})=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4.68)

Then we have hx​uk​y=0subscriptβ„Žπ‘₯subscriptπ‘’π‘˜π‘¦0h_{x}u_{ky}=0 by comparing the coefficients of zksuperscriptπ‘§π‘˜z^{k} of equation (4.68) for 2≀k≀dβˆ’22π‘˜π‘‘22\leq k\leq d-2. Thus, we have hx=0subscriptβ„Žπ‘₯0h_{x}=0 or u(dβˆ’2)​y=β‹―=u2​y=0subscript𝑒𝑑2𝑦⋯subscript𝑒2𝑦0u_{(d-2)y}=\cdots=u_{2y}=0. If u(dβˆ’2)​y=β‹―=u2​y=0subscript𝑒𝑑2𝑦⋯subscript𝑒2𝑦0u_{(d-2)y}=\cdots=u_{2y}=0, then it follows from equation (4.63) that hx=0subscriptβ„Žπ‘₯0h_{x}=0. If hx=0subscriptβ„Žπ‘₯0h_{x}=0, then it follows from equation (4.68) that hy=0subscriptβ„Žπ‘¦0h_{y}=0 or u1​x=u0​x=0subscript𝑒1π‘₯subscript𝑒0π‘₯0u_{1x}=u_{0x}=0. If u1​x=u0​x=0subscript𝑒1π‘₯subscript𝑒0π‘₯0u_{1x}=u_{0x}=0, then it follows from equation (4.63) and (4.62) that v1​x=0subscript𝑣1π‘₯0v_{1x}=0 or u(dβˆ’2)​y=β‹―=u2​y=0subscript𝑒𝑑2𝑦⋯subscript𝑒2𝑦0u_{(d-2)y}=\cdots=u_{2y}=0. If v1​x=0subscript𝑣1π‘₯0v_{1x}=0, then it reduces to Case II(b). If u(dβˆ’2)​y=β‹―=u2​y=0subscript𝑒𝑑2𝑦⋯subscript𝑒2𝑦0u_{(d-2)y}=\cdots=u_{2y}=0, then it follows from equation (4.64) that u1​y=0subscript𝑒1𝑦0u_{1y}=0. It follows from equation (4.65) that u0​y=0subscript𝑒0𝑦0u_{0y}=0. Then it follows from equation (4.66) that hy=0subscriptβ„Žπ‘¦0h_{y}=0. Thus, we have h=0β„Ž0h=0 because h​(0,0)=0β„Ž000h(0,0)=0. Therefore, u,v,hπ‘’π‘£β„Žu,v,h are linearly dependent. ∎

Corollary 4.2.

Let H=(u​(x,y,z),v​(x,y,z),h​(x,y))𝐻𝑒π‘₯𝑦𝑧𝑣π‘₯π‘¦π‘§β„Žπ‘₯𝑦H=(u(x,y,z),v(x,y,z),h(x,y)) be a polynomial map with degz⁑v​(x,y,z)≀1subscriptdegree𝑧𝑣π‘₯𝑦𝑧1\deg_{z}v(x,y,z)\leq 1. Assume that H​(0)=0𝐻00H(0)=0 and the components of H𝐻H are linearly independent over 𝐊𝐊{\bf K}. If J​H𝐽𝐻JH is nilpotent, then
(1)1(1) H𝐻H has the form of Theorem 2.2 in the case degz⁑u=0subscriptdegree𝑧𝑒0\deg_{z}u=0 and (degy⁑u,degy⁑h)≀3subscriptdegree𝑦𝑒subscriptdegreeπ‘¦β„Ž3(\deg_{y}u,\deg_{y}h)\leq 3.
(2)2(2) H𝐻H has the form of Proposition 3.1 in the case degz⁑v=0subscriptdegree𝑧𝑣0\deg_{z}v=0 and (degx⁑v,degx⁑h)≀3subscriptdegreeπ‘₯𝑣subscriptdegreeπ‘₯β„Ž3(\deg_{x}v,\deg_{x}h)\leq 3.
(3)3(3) there exists T∈GL3⁑(𝐊)𝑇subscriptGL3𝐊T\in\operatorname{GL}_{3}({\bf K}) such that Tβˆ’1​H​Tsuperscript𝑇1𝐻𝑇T^{-1}HT has the form of Theorem 2.2 in the case degz⁑uβ‹…degz⁑vβ‰₯1subscriptdegree𝑧⋅𝑒subscriptdegree𝑧𝑣1\deg_{z}u\cdot\deg_{z}v\geq 1 and (deg⁑(u​(x,y,0)βˆ’Ξ»β€‹v​(x,y,0)),deg⁑h​(x,y))≀3degree𝑒π‘₯𝑦0πœ†π‘£π‘₯𝑦0degreeβ„Žπ‘₯𝑦3(\deg(u(x,y,0)-\lambda v(x,y,0)),\deg h(x,y))\leq 3 for some Ξ»βˆˆπŠπœ†πŠ\lambda\in{\bf K}.

Proof.

Since u,v,hπ‘’π‘£β„Žu,v,h are linearly independent, so it follows from Theorem 4.1 that degz⁑u≀1subscriptdegree𝑧𝑒1\deg_{z}u\leq 1. Thus, we have degz⁑u≀1subscriptdegree𝑧𝑒1\deg_{z}u\leq 1 and degz⁑v≀1subscriptdegree𝑧𝑣1\deg_{z}v\leq 1. Then the conclusions (1) and (2) follow from Theorem 2.2 and Proposition 3.1 respectively. The conclusion of (3) follows from the proof of Theorem 3.10 in [12] and Theorem 2.2. ∎

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